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Proof step 1 from proof q-add (conclusion)

⊦ ∃ n. m = n + n ∨ m = (n + n)+1

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ ∃ n. 0 = n + n ∨ 0 = (n + n)+1
Proof step 4:
∃ n. m0 = n + n ∨ m0 = (n + n)+1 ⊦ ∃ n. m0 +1 = n + n ∨ m0 +1 = (n + n)+1

Rule argument

term: m

Additional information

Used simplifier rules: gen: ⊦ ;