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Proof step 16 from proof mod-ls-same-01 (previous step)

abs(j) = 1, 0 < i ⊦ i % j < i ↔ abs(j) = 1 ∨ abs(j) ≤ i, i = 0, j = 0, j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
abs(j) = 1, 0 < i, 0 < i % j, i ≤ i % j ⊦ i = 0, j = 1, j = 0

Rule argument

Additional information

Used simplifier rules: forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)