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Proof step 2 from proof mod-two-05 (previous step)

1 < i ⊦ i % 2 = 0, i % 2 = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mod-two-06: ⊦ 0 < i → (i % 2 = 1 ↔ i % 2 ≠ 0);%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1