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Proof step 13 from proof eq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i0 ^ i2 * i0 = i0 ^ i * i0, ~2 < i, ~2 < i2, 1 < i0, 2 ≤ i0, 0 ≤ i2, ~1 ≤ i2, 0 ≤ i, ~1 ≤ i, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i1 ∧ 0 ≤ j → i1 = j ⊦ i2 = i

Rule argument

Additional information

Used simplifier rules: numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ i +1 = i + 1; (s)