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Proof step 7 from proof eq-pos (previous step)

i0 ^ i -1 * i0 = i0 ^ j, 1 < i0, 0 ≤ i, 0 ≤ j, 2 ≤ i0, 0 < i, ∀ i1, i0, j. i→n(i1) < i→n(i) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i = j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
i = i2 +1, i0 ^ i2 * i0 = i0 ^ j, 1 < i0, 0 ≤ i, 0 ≤ j, 2 ≤ i0, 0 < i, ∀ i1, i0, j. i→n(i1) < i→n(i) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i = j

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [i, i2] flags:

Additional information

Used simplifier rules: ⊦ i = j -1 ↔ j = i +1; (elim) Used terms: i, i2