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Proof step 3 from proof one-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
j = i +1, 0 < j, ∀ j0. i→n(j0) < i→n(j) → true → 0 < j0 → 1 ^ j0 = 1 ⊦ 1 ^ i = 1

Rule argument

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

Additional information

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