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Proof step 2 from proof absls-05 (previous step)

abs(i - 1) < abs(i) ⊦ 0 < i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%mone-06: ⊦ ~1 < i ↔ 0 ≤ i;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%abs-08: ⊦ i < 0 → abs(i) = ~ i;%Minus-def: ⊦ ~ i = 0 - i;%abs-05: ⊦ i < 0 → abs(i + 1) = abs(i) - 1;%sub-sub: ⊦ i - j - i0 = i - j + i0;%num-subadd: ⊦ (* i) - j + (* i0) = i - i0 - j;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%mone-01: ⊦ i + ~1 = i - 1;%sub-add-09: ⊦ i - j + i = ~ j;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i, 1, i0