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Proof step 9 from proof odd-rec (previous step)

0 ≤ i0 ⊦ (∃ j. 2 * j = i0 + 1) ↔ (∃ j. i0 = j + j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult