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Proof step 8 from proof numls-07 (previous step)

j < 0, 0 < i1, 1 < i1, (1 - i1) * j ≤ 0 ⊦ j = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [1, i1, i]

Additional information

Used simplifier rules: zero-le-01: ⊦ i < 0 → (i * j ≤ 0 ↔ 0 ≤ j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ 0 < i ∧ 0 ≤ j → (1 = i + j ↔ i = 1 ∧ j = 0); (ws)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: 1, i1, i