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Proof step 3 from proof square-even (previous step)

i * i + i = 2 * j + 1 ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ (∃ j. i = 2 * j) ∨ (∃ j. i = 2 * j + 1)
Proof step 5:
i * i + i = 2 * j + 1, (∃ j. i = 2 * j) ∨ (∃ j. i = 2 * j + 1) ⊦

Rule argument

spec: <toplevel> name: ex-two seq: ⊦ (∃ j. i = 2 * j) ∨ (∃ j. i = 2 * j + 1) subst: [i] → [i]

Additional information

Used terms: i