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Proof step 2 from proof oex (previous step)

⊦ ∃ m. n = 2 * m + 1, ∃ m. 2 * m = n

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ ∃ n. m = n + n ∨ m = (n + n)+1
Proof step 4:
∃ n0. n = n0 + n0 ∨ n = (n0 + n0)+1 ⊦ ∃ m. n = 2 * m + 1, ∃ m. 2 * m = n

Rule argument

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

Additional information

Used terms: n