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

⊦ ∃ j. i = 2 * j, ∃ j. i = 2 * j + 1

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ ∃ i. i + i = j ∨ i + i + 1 = j
Proof step 4:
∃ i0. i0 + i0 = i ∨ i0 + i0 + 1 = i ⊦ ∃ j. i = 2 * j, ∃ j. i = 2 * j + 1

Rule argument

spec: int-basic2 inst: name: ex-oddeven seq: ⊦ ∃ i. i + i = j ∨ i + i + 1 = j subst: [j] → [i] flags:

Additional information

Used terms: i