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Proof step 1 from proof modex-odd (conclusion)

⊦ i % 2 = 1 ↔ (∃ j. i = 2 * j + 1 ∧ 0 ≤ j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: mod-two-04: ⊦ 0 ≤ i → (i * 2 + 1) % 2 = 1;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult