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Proof step 20 from proof equal-minus-one (previous step)

i + i ^ (j - 1) = 0, abs(i ^ (j - 1)) = 1, 0 < j ⊦ i = 0, i = ~1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
i + i ^ i0 = 0, abs(i ^ i0) = 1, j = i0 + 1, 0 < j ⊦ i = ~1, i = 0

Rule argument

spec: int-abs name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i0]

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i0