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Proof step 5 from proof zero (previous step)

0 < j, i ^ j -1 * i = 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → true → 0 < j0 → (i ^ j0 = 0 ↔ i = 0) ⊦ i = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
j = i0 +1, 0 < j, i ^ i0 * i = 0, 0 < j, ∀ j0, i. i→n(j0) < i→n(j) → true → 0 < j0 → (i ^ j0 = 0 ↔ i = 0) ⊦ i = 0

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [j, i0] flags:

Additional information

Used simplifier rules: ⊦ i = j -1 ↔ j = i +1; (elim) Used terms: j, i0