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Proof step 2 from proof two-odd (previous step)

i % 2 = 0, 0 < i ⊦ i = 2 * ((i - 1) / 2 + 1)

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: exeven seq: i % 2 = 0 ⊦ ∃ j. i = 2 * j subst: [i] → [i] flags: current, []

Additional information

Used simplifier rules: mod-two-02: ⊦ (i * 2) % 2 = 0;%two-mult-01: ⊦ 0 < i → (2 * i - 1) / 2 = i - 1;%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%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%Sub-def: ⊦ i + (j - i) = 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 Used terms: i