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Proof step 18 from proof isSquare-mult (previous step)

j1 * j1 = abs(j), j0 * j0 = j * i * i, abs(j0) = abs(j1) * abs(i) ⊦ i = 0, ∃ j0. j0 * j0 = j, 0 < j0 * j0 ∧ j < 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%square-lez-01: ⊦ i * i ≤ 0 ↔ i = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%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%add-zero: ⊦ i + 0 = i; from specification int-basic2%square-zero: ⊦ i * i = 0 ↔ i = 0; from specification int-mult Used terms: j