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Proof step 20 from proof square-eq (previous step)

¬ ¬ i < 0, abs(i) = abs(j) ⊦ i * i = j * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i + abs(j) = 0, i < 0 ⊦ i * i = j * j, j = 0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs