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Proof step 7 from proof modex-odd-neg-01 (previous step)

2 * j = i + 1, j ≤ 0 ⊦ i = 2 * (j - 1) + 1 ∧ j - 1 < 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
j = i0 + 1, 2 * j = i + 1, j ≤ 0 ⊦ i = 2 * i0 + 1 ∧ i0 < 0

Rule argument

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

Additional information

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