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Proof step 15 from proof div-mult-two-nls-01 (previous step)

(2 * i0) * j + 2 * j0 = i1 * j + j1, j0 < j, 1 + 2 * i0 < i1, 0 < j, j1 < j, 0 ≤ i0, 0 ≤ j0, 0 ≤ j1, 0 ≤ i1 ⊦ 0 ≤ 2 * i0 ∧ 0 ≤ i1 ∧ 0 ≤ 2 * j0 ∧ 0 ≤ j1 ∧ 2 * j0 < j ∧ j1 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: nummult-ls-02: ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ 0 < i → (i < i * j ↔ 1 < j); (s)% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j + i * j0 = i * (j + j0); (s)