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Proof step 13 from proof odd-rec (previous step)

2 * (i1 + 1) = i0 + 1, 0 ≤ i0 ⊦ i0 = 2 * i1 + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: numadd-01 seq: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0 subst: [j0, i0, j] → [1, i0, 2 * (i1 + 1)] flags: rewrite

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult Used terms: 1, i0, 2 * (i1 + 1)