S01 · implies
Before
A=\{a,b\},\ B=\{1,2,3\}After
|A|=2,\quad |B|=3
Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
Both carriers are complete with distinct entries, so their counts contain no duplicates.
derivation-mko-cardinality-comparison-01 · ocme-new142-reader-adoption-20261021
There is an injection A to B with |A|=2<3=|B|; a reverse surjection is not injective.
Explicit finite sets and full maps only; no infinite-choice claim.
calcG1 — A={a,b}, B={1,2,3}; f(a)=1,f(b)=2; g(1)=a,g(2)=b,g(3)=a.f,g — maps with opposite directionsA,B,C — complete finite setscalc · prerequisite — Assume finite counting and the definitions of injection and surjection.mko_definition · cited — This slice is limited to complete finite sets. An injection A→B certifies |A|≤|B|, a surjection certifies |A|≥|B|, and a bijection certifies equality. If |A|>|B| there is no injection. Reverse comparison from a surjection for infinite sets requires additional choice or section obligations and is outside this slice. · mko-cardinality-comparisonBefore
A=\{a,b\},\ B=\{1,2,3\}After
|A|=2,\quad |B|=3
Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
Both carriers are complete with distinct entries, so their counts contain no duplicates.
Before
f:A\to B
After
\operatorname{graph}(f)=\{(a,1),(b,2)\}Conditions: none
Rule: calc · Uses: G1, S01
Leaf basis: calc
Each source element has exactly one image in B, so f is total.
Before
f(a)=1,\ f(b)=2
After
f(a)\ne f(b)\Rightarrow f\text{ injective}Conditions: none
Rule: calc · Uses: G1, S02
Leaf basis: calc
A has only two elements, and the only distinct pair has distinct images, exhausting collision checks.
Before
f(A)=\{1,2\}After
3\notin f(A),\quad f\text{ not surjective},\quad 2\le3Conditions: none
Rule: mko_definition · Uses: G1, S01, S03
Leaf basis: mko_definition
Missing 3 refutes surjectivity, while the two distinct images certify |A|≤|B|.
Before
g:B\to A
After
g^{-1}(\{a\})=\{1,3\},\quad g^{-1}(\{b\})=\{2\}Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
The full reverse map gives two nonempty fibers accounting for all of B.
Before
g(B)=A,\quad g(1)=g(3)
After
g\text{ surjective but not injective};\quad3\ge2Conditions: none
Rule: mko_definition · Uses: G1, S05
Leaf basis: mko_definition
Both a and b have preimages, so g is onto; distinct 1 and 3 have the same image, so g is not injective.
Before
h:B\to A\text{ assumed injective}After
h(1),h(2),h(3)\text{ distinct in a two-element set: impossible}Conditions: none
Rule: calc · Uses: G1, S01
Leaf basis: calc
An injection would require three distinct elements in explicitly two-element A, which is impossible.
Before
C=\{u,v\},\quad b(a)=u,\ b(b)=vAfter
b:A\to C\text{ bijective};\quad |A|=|C|=2Conditions: none
Rule: mko_definition · Uses: G1, S01
Leaf basis: mko_definition
A separate full map is both injective and onto u,v, certifying equality of cardinalities.
No substeps in this guide.
S08 · premise_audit · G1, S01 — The equality has an actual bijection witness and does not mislabel the original f as onto.S07 · premise_audit · G1, S01 — This excludes every h, not merely g, using the complete carrier sizes.S06 · premise_audit · G1, S05 — Audit onto and one-to-one separately; the reverse surjection gives the finite reverse inequality only.S05 · premise_audit · G1 — A fiber is a source subset; inputs 1 and 3 cannot be counted as one element.S04 · premise_audit · G1, S01, S03 — Do not upgrade injection to bijection; the missing image is certified by the full source table.S03 · premise_audit · G1, S02 — 1 and 2 are distinct in B; injectivity does not require hitting 3.S02 · premise_audit · G1, S01 — Checking filled entries alone would miss an omitted source element.S01 · premise_audit · G1 — Fix the sets first; changing a codomain may change surjectivity.f needs two entries and g three; source lengths cannot be interchanged.
An injection A to B gives ≤, whereas a surjection in that direction gives ≥.
reject — 3 has no preimage.reject — g(1)=g(3)=a with 1 distinct from 3.reject — The certificates concern A,B,C only and handle neither infinite sets nor choice principles.There is an injection A to B with |A|=2<3=|B|; a reverse surjection is not injective.
1 declared checks for this guide; only the listed expressions are covered.
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