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Proof step 3 from proof mod-ls-zero (previous step)

0 < j ⊦ i % j < 0 ↔ i < 0 ∧ i % j ≠ 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j;%mod-zero-01: ⊦ 0 < j ∧ 0 ≤ i → ¬ i % j < 0;%mod-grz: ⊦ 0 < j ∧ i < 0 → ¬ 0 < i % j;%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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