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Proof step 23 from proof square-eq (previous step)

~ abs(j) + abs(j) = 0, ~ abs(j) < 0 ⊦ ~ abs(j) * ~ abs(j) = j * j, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
⊦ abs(j) * abs(j) = j * j, j = 0

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ ~ ~ i = i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i + i = 0; (s)