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Proof step 2 from proof square-even (previous step)

i * i + i = 2 * j + 1 ⊦

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: int-mult inst: name: ex-two seq: ⊦ (∃ j. i = 2 * j) ∨ (∃ j. i = 2 * j + 1) subst: [i] → [i] flags: current, []

Additional information

Used terms: i