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Proof step 29 from proof gcd-mult-pos (previous step)

j * i1 ≠ 0, 0 < i1, 0 < i, 0 ≤ j, j ≤ i, 0 ≤ i, ∀ j0, i0, i2. i→n(j0 * i0 + i2 * i0) < i→n(j * i1 + i * i1) → 0 < i0 ∧ 0 < i2 ∧ 0 ≤ j0 → gcd(i2 * i0, j0 * i0) = gcd(i2, j0) * i0 ⊦ gcd(abs(j * i1), abs(i * i1) % abs(j * i1)) = gcd(i, j) * i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%mod-mult-02: 0 ≠ j, i0 ≠ 0 ⊦ (i0 * i) % (i0 * j) = i % j * i0; from specification int-div%zero-left: ⊦ 0 * i = 0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult