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Proof step 20 from proof abs-ls-same-01 (previous step)

j < 0, i ≤ abs(i) * j ⊦ i = 0, j = ~1

User applied rule cut formula

And obtained the following new goals

Proof step 21:
i + i * j = i * (1 + j), j < 0, i ≤ abs(i) * j ⊦ i = 0, j = ~1

Rule argument

fma: i + i * j = i * (1 + j)

Additional information

Used simplifier rules: eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%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 + i * j = i * (1 + j)