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Proof step 2 from proof mult-sub-less (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: ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0);%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 ↔ false); (as)% ⊦ i + j ≤ i ↔ j ≤ 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i - j < i0 ↔ i < j + i0; (s)