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Proof step 2 from proof abs-pred (previous step)

i→n(i - 1) < i→n(i), i < 0 ⊦ i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < i → i→n(i) = i→n(i - 1) + 1
Proof step 4:
i < 0, i→n(i - 1) < i→n(i) ⊦ i = 0, 0 < i

Rule argument

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

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i; (s) Used terms: i