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Proof step 6 from proof odd-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i0 ⊦ (∃ j. 2 * j = i0 + 1) ↔ odd(i0)

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%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