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Proof step 2 from proof mult-sub-lesseq (previous step)

0 < i, 0 < j, 0 < i1 ⊦ (i - j) * i1 ≤ i * i1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-pos: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%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; (s)% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i - j ≤ i0 ↔ i ≤ j + i0; (s)