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Proof step 10 from proof div-mod-01 (previous step)

i0 = i3 * j + j0, 0 < j, j0 < j, 0 ≤ i2 * j + j0, 0 ≤ i2, i0 ≤ i2 * j + j0, 0 ≤ i0, 0 ≤ j0, 0 ≤ i3 ⊦ ∃ i1. i2 * j + j0 = j * i1 + i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 < j, j0 < j, i3 ≤ i2, 0 ≤ j0, 0 ≤ i3, 0 ≤ i2 ⊦ ∃ i. i2 = i3 + i

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); (s)% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ i * j + i * j0 = i * (j + j0); (s)% ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0); (s)% ⊦ i + j ≤ i0 + j ↔ i ≤ i0; (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 ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); (s)