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Proof step 21 from proof equal-minus-one (previous step)

abs(i ^ i0) = 1, i + i ^ i0 = 0, 0 ≤ i0 ⊦ i = 0, i = ~1

User applied rule cut formula

And obtained the following new goals

Proof step 22:
abs(i ^ i0) = 1, i + i ^ i0 = 0, 0 ≤ i0 ⊦ i0 = 0, i = 0, i = ~1

Rule argument

fma: i0 = 0

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;%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: i0 = 0