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Proof step 18 from proof equal-minus-one (previous step)

0 < j, i ^ j -1 * i = ~1, 0 < j ⊦ i = 0, i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
abs(i ^ (j - 1)) = 1, i + i ^ (j - 1) = 0, 0 < j ⊦ i = 0, i = ~1

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%eq-one-neg: ⊦ i * j = ~1 ↔ abs(i) = 1 ∧ i = ~ j; from specification int-mult%sub-pred: ⊦ i -1 = i - 1; from specification int-abs