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Proof step 1 from proof ins-01 (conclusion)

⊦ sort(ins≤(a, x) + y) = ins≤(a, sort(x + y))

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ a, y. sort(ins≤(a, x0) + y) = ins≤(a, sort(x0 + y)) ⊦ sort(ins≤(a, a0 + x0) + y) = ins≤(a, sort((a0 + x0) + y))

Rule argument

indvar: x

Additional information

Used simplifier rules: rec: ⊦ sort(a ' + x) = ins≤(a, sort(x));%list freely generated by [], +;;%ins-e: ⊦ ins≤(a, []) = a '; from specification olist%a: ⊦ (x + y) + z = x + y + z; from specification list%Nil: ⊦ [] + x = x; from specification list%eq: ⊦ ins≤(a, x) = ins≤(a, y) ↔ x = y; from specification olist