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Proof step 3 from proof eq-one-neg (previous step)

i * j = ~1 ⊦ abs(j) = 1 ∧ j + i = 0

User applied rule cut formula

And obtained the following new goals

Proof step 4:
~(i * j) = 1, i * j = ~1 ⊦ abs(j) = 1 ∧ j + i = 0

Rule argument

fma: ~(i * j) = 1

Additional information

Used simplifier rules: negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: ~(i * j) = 1