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Proof step 6 from proof ex-two (previous step)

i0 + i0 = i ∨ i0 + i0 + 1 = i ⊦ i = 2 * i0, ∃ j. i = 2 * j + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numeq-03: ⊦ (* i2) * i = i + j0 ↔ (i2 - 1) * i = j0;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2