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Proof step 9 from proof modeqp (previous step)

i1 = i +1, i * j + j0 = i0 * j + j1, 0 ≤ j0, j0 < j, 0 ≤ j1, j1 < j, 0 < j, ∀ j, j0, i0, j1. i1 * j + j0 = i0 * j + j1 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ j1 ∧ j1 < j ∧ 0 < j → i1 = i0 ∧ j0 = j1 ⊦ i0 + 1 = i1 ∧ j0 = j1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i * j + j0 = i0 * j + j1, j0 < j, j1 < j, 0 < j, 0 ≤ j0, 0 ≤ j1, ∀ j2, j3, i1, j4. (i * j2 + j2) + j3 = i1 * j2 + j4 ∧ 0 ≤ j3 ∧ j3 < j2 ∧ 0 ≤ j4 ∧ j4 < j2 ∧ 0 < j2 → i + 1 = i1 ∧ j3 = j4 ⊦ i0 = i ∧ j0 = j1

Rule argument

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2