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Proof step 6 from proof eq-pos (previous step)

0 < i, i0 ^ i -1 * i0 = i0 ^ j, 2 ≤ i0, 0 ≤ j, 0 ≤ i, 1 < i0, ∀ 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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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

Rule argument

Additional information

Used simplifier rules: leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs