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Proof step 10 from proof noeth-zero-ls (previous step)

0 < (i - 1) * j, i * j < (i - 1) * j, 0 < i, 0 < i1, i * j < i, i * j < j, i * j ≤ 0, 0 < j, i ≤ i1, i * j < i1, ∀ i0, j0. abs(i0) < i1 ∧ i0 ≠ 0 ∧ 0 < j0 → 0 < abs(i0) * j0 ⊦ i = 1, j = 0

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, 1, i2] 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;%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; 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%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i2