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Proof step 3 from proof numls-03 (previous step)
| ⊦ i2 * i + j < i + j0 ↔ (i2 - 1) * i + j < j0 |
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] → [i2, 1, i0]
flags: Additional information
Used simplifier rules:
dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2% ⊦ i0 = i - j ↔ i = i0 + j; (elim)
Used terms: i2, 1, i0