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Proof step 2 from proof mod-le-num (previous step)

j < j0, 0 < j ⊦ i % j ≤ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
j < i % j, j0 < i % j, j < j0, 0 < j, 0 < j0, 1 < j0, j + 1 < i % j, 1 < i % j, 0 < i % j, 2 < i % j ⊦

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs