Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-02 (claude-opus-5)
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Injectivity of the successor is independent of the other Peano axioms

Statement refuted

Refuted claim: (P2) injectivity of σ follows from (P1) 0 is not a successor together with (P3) induction. The witness is the three-element system N={0,1,2} with σ(0)=1, σ(1)=2, σ(2)=2.

Facts & Assumptions

Given: N={0,1,2}, base point 0, σ(0)=1, σ(1)=2, σ(2)=2.

[L1]

The Peano axioms (Peano system).

Counterexample

technique · direct
1.1

P1 holds: the values of σ are σ(0)=1, σ(1)=2, σ(2)=2, none of which is 0, so σ(x)≠0 for all x.

givenL1
1.2

P3 holds: if S⊆N with 0∈S and S closed under σ, then 0∈S forces 1=σ(0)∈S, then 2=σ(1)∈S, so S={0,1,2}=N; thus induction holds.

givenL1
1.3

P2 fails: σ(1)=2=σ(2) but 1≠2, so σ is not injective.

givenL1
2.1

So (N,0,σ) satisfies P1 and P3 but not P2, refuting the claim; note the failure permits the pathology σ(2)=2 (a number that is its own successor) and a finite model in which distinct numerals collapse.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources