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Proof step 18 from proof mod-ls-same-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
0 < i ⊦ abs(j) = 1, j = 1, j = 0, i = 0, i % j < i ↔ abs(j) ≤ i

Rule argument

Additional information