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Proof step 8 from proof auxexdivmod-01 (previous step)

0 < j, 0 < i, 0 ≤ i, j ≤ 1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < i0 * j + j ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ i = 1 → i * j = j;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ ¬ i < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)