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Proof step 2 from proof mod-lsz-01 (previous step)

⊦ i % j < 0 ↔ i < 0 ∧ i % j ≠ 0, j = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: mod-zero-06: ⊦ 0 ≤ i ∧ j ≠ 0 → ¬ i % j < 0;%mod-grz-01: ⊦ i < 0 ∧ j ≠ 0 → ¬ 0 < i % j;%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%irref: ⊦ ¬ i < i; from specification int-basic1