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Proof step 8 from proof eq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
i0 ^ i2 * i0 = i0 ^ j, ~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

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%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2