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Proof step 16 from proof mod-ls-same-01 (previous step)

abs(j) = 1, 0 < i, i ≤ i % j, 0 < i % j ⊦ j = 1, j = 0, i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j] flags: rewrite

Additional information

Used simplifier rules: modnum-neg-01: ⊦ j < 0 → i % (* j) = i % ~ j;%mod-one: ⊦ i % 1 = 0;%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%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%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs Used terms: j