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Proof step 10 from proof prime-5 (previous step)

5 % i = 0, i * i ≤ 5, 0 < i, 1 < i ⊦ 3 ≤ i, i = 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
5 % i = 0, i * i ≤ 5, 0 < i, 1 < i, i < 3 ⊦ i = 2

Rule argument

Additional information

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