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Proof step 1 from proof exodd (conclusion)

i % 2 = 1, 0 ≤ i ⊦ ∃ j. 0 ≤ j ∧ i = 2 * j + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i % 2 = 1, 0 ≤ i ⊦ ∃ j. 0 ≤ j ∧ i = 2 * j + 1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs