S01 · implies
Before
f=(a,b,a,b,a)
After
\forall x\in A:\ \exists!y\in B:\ f(x)=y
Conditions: none
Rule: work · Uses: G1, work
Leaf basis: work
Each of the five domain elements has one legal image, proving totality.
derivation-mko-pigeonhole-principle-01 · ocme-new142-reader-adoption-20261021
Use all fibres to prove a collision and the generalized lower bound.
A finite total map with nonempty codomain; no sampling inference about general maps.
workG1 — A={0,1,2,3,4}, B={a,b}, with complete f table(a,b,a,b,a).ℓ_b — load of box bm,k — domain and codomain sizesmko_definition · cited — For complete finite A,B and total f:A→B, |A|>|B| implies distinct a1,a2 with f(a1)=f(a2). The reverse size condition does not generally certify a given map injective. The generalized maximum-load bound is ceil(|A|/|B|), with a nonempty codomain. · mko-pigeonhole-principlework · prerequisite — Fibres of a finite total map form a partition; loads sum to the domain size; ceil(r)-1<r.Before
f=(a,b,a,b,a)
After
\forall x\in A:\ \exists!y\in B:\ f(x)=y
Conditions: none
Rule: work · Uses: G1, work
Leaf basis: work
Each of the five domain elements has one legal image, proving totality.
Before
|A|=5,\quad|B|=2
After
|A|>|B|
Conditions: none
Rule: work · Uses: G1, S01, work
Leaf basis: work
The finite sizes satisfy the strict inequality.
Before
f^{-1}(a)After
\{0,2,4\},\quad\ell_a=3Conditions: none
Rule: work · Uses: G1, S02, work
Leaf basis: work
List the complete fibre over a and its load.
Before
f^{-1}(b)After
\{1,3\},\quad\ell_b=2Conditions: none
Rule: work · Uses: G1, S03, work
Leaf basis: work
Complete the other fibre; both are disjoint and cover A.
Before
0\ne2,\quad f(0)=a=f(2)
After
f\text{ is not injective}Conditions: none
Rule: work · Uses: G1, S04, work
Leaf basis: work
Give an actual collision witness with two distinct domain elements.
Before
\lceil|A|/|B|\rceil=\lceil5/2\rceil
After
\max(\ell_a,\ell_b)=3\ge3
Conditions: none
Rule: work · Uses: G1, S05, work
Leaf basis: work
The generalized ceiling bound is 3 and is attained here.
Before
\ell_b\le q-1\text{ for every box},\quad q=\lceil m/k\rceilAfter
m=\sum_b\ell_b\le k(q-1)<m\quad(k\ge1)
Conditions: none
Rule: work · Uses: G1, S06, work
Leaf basis: work
If every load were below the ceiling, their sum would contradict the domain size m; therefore at least one load is at least ceil(m/k).
Before
A'=\{0,1\},\ B'=\{a,b\},\ g=(a,a)After
|A'|\le|B'|\quad\text{but }g\text{ not injective}Conditions: none
Rule: work · Uses: G1, S07, work
Leaf basis: work
A complete counterexample refutes the erroneous converse size condition.
No substeps in this guide.
S08 · premise_audit · G1, S07, work — A smaller domain does not force an arbitrary map to be injective.S07 · premise_audit · G1, S06, work — The ceiling definition gives q-1<m/k; multiply by positive k.S06 · premise_audit · G1, S05, work — A nonempty codomain is required for this quotient.S05 · premise_audit · G1, S04, work — f(0)=f(0) is not a witness using distinct elements.S04 · premise_audit · G1, S03, work — The load sum3+2=5 cross-checks completeness.S03 · premise_audit · G1, S02, work — 0,2,4 are three distinct preimages.S02 · premise_audit · G1, S01, work — Equal sizes would not trigger this collision guarantee.S01 · premise_audit · G1, work — The entry for4 cannot be omitted.List all five entries with values in B.
Round5/2 upward to 3, not truncation.
reject — The preimages must be distinct.reject — g=(a,a) is a complete counterexample.reject — The theorem's bound is the ceiling3, not the floor2.0 and 2 collide and maximum load3=ceil(5/2); the reverse size condition does not ensure injectivity.
2 declared checks for this guide; only the listed expressions are covered.
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