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Proof step 21 from proof square-eq (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 22:
i = ~ abs(j), i + abs(j) = 0, i < 0 ⊦ i * i = j * j, j = 0

Rule argument

fma: i = ~ abs(j)

Additional information

Used simplifier rules: negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i = ~ abs(j)