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Proof step 8 from proof modzero-08 (previous step)

0 < i, abs(~(i * j)) < j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < i, i * abs(j) < j ⊦

Rule argument

Additional information

Used simplifier rules: abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ 0 < j → abs(j) = j; (s)% ⊦ abs(~ i) = abs(i); (s)