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Proof step 18 from proof mod-ls-same-02 (previous step)

i0 * j + j0 < 0, j < j0, j < 0, j * (i0 + 1) + j0 ≤ 0, j0 ≤ 0, 0 ≤ i0 ⊦ 0 < i0 ↔ ~1 = j ∨ ¬ 0 < j * (i0 - 1) + j0, 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] → [i0, 1, i]

Additional information

Used simplifier rules: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 ≤ i ∧ j < 0 ∧ j0 ≤ 0 → ¬ 0 < i * j + j0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ 1 * i = i; (s)% ⊦ i * 0 = 0; (s)% ⊦ i + 0 = i; (s)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i * ~1 = ~ i; (s)% ⊦ i + ~ j = i - j; (s)%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ i - j ≤ i0 ↔ i ≤ j + i0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0, 1, i