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Proof step 5 from proof rax-03 (previous step)

i0 < 0 ⊦ i→n(~ i0) = i→n(i0 + 1) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < i → i→n(i) = i→n(i - 1) + 1
Proof step 7:
i0 < 0, 0 < ~ i0 ⊦ i→n(~ i0 - 1) + 1 = i→n(i0 + 1) + 1

Rule argument

spec: <toplevel> name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [~ i0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: ~ i0