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Proof step 19 from proof abs-ls-same-01 (previous step)

j < 0 ⊦ abs(i) * j < i ↔ ¬ i ≤ 0 ∨ i ≠ 0 ∧ j ≠ ~1, j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
j < 0, i ≤ abs(i) * j ⊦ i = 0, j = ~1

Rule argument

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ i < j ↔ ¬ j ≤ i; (ss)