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Proof step 6 from proof prime-5 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
5 % i = 0, 3 ≤ i, i * i ≤ 5, 0 < i, 0 ≤ i ⊦ i = 2

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%numlsledel-01: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j; from specification int-abs