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Proof step 1 from proof mod-ls-zero (conclusion)
0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0
Special step normalize
And obtained the following new goals
Proof step 2
:
0 < j ⊦ i % j < 0 ↔ i < 0 ∧ i % j ≠ 0