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Proof step 4 from proof not-odd-ex (previous step)

⊦ (∃ j. i = j + j + 1) ↔ ¬(∃ j. i = 2 * j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ ∃ j. i = 2 * j + 1, ∃ j. i = 2 * j

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 2 * i ≠ 2 * j + 1; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + i = 2 * i; (s)