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

⊦ ∃ n. m = 2 * n, ∃ n. m = 2 * n + 1

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ ∃ m. m + m = n ∨ m + m + 1 = n
Proof step 4:
∃ m0. m0 + m0 = m ∨ m0 + m0 + 1 = m ⊦ ∃ n. m = 2 * n, ∃ n. m = 2 * n + 1

Rule argument

spec: nat-basic2 inst: name: ex-oddeven seq: ⊦ ∃ m. m + m = n ∨ m + m + 1 = n subst: [n] → [m] flags:

Additional information

Used terms: m