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Proof step 11 from proof eq-pos (previous step)

0 < j, i0 ^ i2 * i0 = i0 ^ j -1 * i0, ~2 < i2, 2 ≤ i0, 0 ≤ j, 1 < i0, 0 ≤ i2, ~1 ≤ i2, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i2 + 1 = j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
j = i +1, 0 < j, i0 ^ i2 * i0 = i0 ^ i * i0, ~2 < i2, 2 ≤ i0, 0 ≤ j, 1 < i0, 0 ≤ i2, ~1 ≤ i2, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i2 + 1 = j

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