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Proof step 9 from proof eq-one-neg (previous step)

j + i = 0, abs(j) = 1 ⊦ i * j = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [j] flags: <all paths>

Additional information

Used simplifier rules: minus-one-01: ⊦ i * ~1 = ~ i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i = j → i - j = 0; (s) Used terms: j