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Proof step 2 from proof sub-ls-01 (previous step)

(i - j) / i0 < i0, (i - j) / i0 < 0, 0 < i0, j ≤ i, (i - j) / i0 + 1 < i0 ⊦

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, j, i1] flags:

Additional information

Used simplifier rules: div-zero-nls: ⊦ 0 ≤ i ∧ 0 < j → ¬ i / j < 0;%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; 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% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, j, i1