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Proof step 13 from proof modex (previous step)

i % j = 0 ⊦ 0 < j, j = 0, j < 0 ∧ ¬ i < 0, 0 < j ∧ ¬ i < 0, ∃ j0. i = j0 * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-pos-not-neg: ⊦ 0 < i1 ∧ 0 < i % i1 → ¬ i < 0;%mod-ls-same-02: j ≠ 0 ⊦ i < i % j ↔ i < 0 ∧ (abs(j) = 1 ∨ abs(j) ≤ abs(i));%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1