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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261004-advance-20261005-r02-surjective-foundations · 2026-10-04 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
The declared total function f:A→B has A=∅, B=∅, and table ∅. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f surjective onto the declared codomain B?
Yes
Check a preimage for every target in B, giving Yes. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. Is f surjective onto the declared codomain B?
No
Check a preimage for every target in B, giving No. Multiple preimages are allowed, and the attained range is not automatically the declared codomain.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-table-property · candidate
The declared total function f:A→B has A=∅, B=∅, and table ∅. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{5}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {5}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{-3, -2}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-3, -2}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{7}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {7}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{-4, -3}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-4, -3}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{9}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {9}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{-5, -4, -3}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-5, -4, -3}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{11}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {11}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{-6, -5, -4}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-6, -5, -4}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{13}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {13}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{-7, -6, -5, -4}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-7, -6, -5, -4}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{15}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {15}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
∅
Collect the attained outputs and remove them from the declared codomain. The uncovered set is ∅; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. List the uncovered targets B∖f[A] in increasing order, using ∅ for the empty set.
{-8, -7, -6, -5}
Collect the attained outputs and remove them from the declared codomain. The uncovered set is {-8, -7, -6, -5}; any missing target defeats surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-missing-targets · candidate
The declared total function f:A→B has A=∅, B={0}, and table ∅. For t=0∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. For t=1∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. For t=5∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. For t=0∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. For t=-2∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. For t=2∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. For t=7∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. For t=1∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. For t=-4∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. For t=1∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-3
Enumerate the full fiber and select its least input if it is nonempty. The answer is -3. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. For t=9∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. For t=2∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. For t=-3∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. For t=3∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-2
Enumerate the full fiber and select its least input if it is nonempty. The answer is -2. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. For t=11∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. For t=3∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. For t=-2∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-2
Enumerate the full fiber and select its least input if it is nonempty. The answer is -2. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. For t=5∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. For t=13∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. For t=4∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. For t=-1∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-5
Enumerate the full fiber and select its least input if it is nonempty. The answer is -5. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. For t=0∈B, find its numerically least permitted preimage, answering none when there is no preimage.
0
Enumerate the full fiber and select its least input if it is nonempty. The answer is 0. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. For t=15∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. For t=5∈B, find its numerically least permitted preimage, answering none when there is no preimage.
-1
Enumerate the full fiber and select its least input if it is nonempty. The answer is -1. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. For t=-8∈B, find its numerically least permitted preimage, answering none when there is no preimage.
none
Enumerate the full fiber and select its least input if it is nonempty. The answer is none. Existence of a preimage does not imply uniqueness.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-first-preimage · candidate
The declared total function f:A→B has A=∅, B=∅, and table ∅. The new codomain B2=∅ contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. The new codomain B2={1, 0, 101} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. The new codomain B2={1, 3} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. The new codomain B2={0, 103} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. The new codomain B2={-1} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. The new codomain B2={1, 2, 0, 105} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. The new codomain B2={1, 3, 5} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. The new codomain B2={0, 1, 107} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. The new codomain B2={-1, -2} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. The new codomain B2={1, 2, 3, 0, 109} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. The new codomain B2={1, 3, 5, 7} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. The new codomain B2={0, 1, 2, 111} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. The new codomain B2={-1, -2} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. The new codomain B2={1, 2, 3, 4, 0, 113} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. The new codomain B2={1, 3, 5, 7, 9} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. The new codomain B2={0, 1, 2, 3, 115} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. The new codomain B2={-1, -2, -3} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. The new codomain B2={1, 2, 3, 4, 5, 0, 117} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. The new codomain B2={1, 3, 5, 7, 9, 11} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. The new codomain B2={0, 1, 2, 3, 4, 119} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. The new codomain B2={-1, -2, -3} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. The new codomain B2={1, 2, 3, 4, 5, 6, 0, 121} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. The new codomain B2={1, 3, 5, 7, 9, 11, 13} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. The new codomain B2={0, 1, 2, 3, 4, 5, 123} contains every attained output. Is the same output table surjective as A→B2?
No
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain No; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. The new codomain B2={-1, -2, -3, -4} contains every attained output. Is the same output table surjective as A→B2?
Yes
Containing the range first makes the new total-function declaration valid. Compare that range with B2 to obtain Yes; no existing output may be discarded to force surjectivity.
Reviewed step-by-step derivation
Surjectivity and codomain coverage · qf-surjective-codomain-change · candidate
Canonical question JSONL · English JSONL · Validation report