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Proof step 5 from proof del1 (previous step)

a ≠ a0, ∀ a. sort(x0 -1l a) = sort(x0) -1l a ⊦ sort(a0 ' + (x0 -1l a)) = ins≤(a0, sort(x0)) -1l a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a. sort(x0 -1l a) = sort(x0) -1l a ⊦ a = a0, ins≤(a0, sort(x0 -1l a)) = ins≤(a0, sort(x0)) -1l a

Rule argument

Additional information

Used simplifier rules: rec: ⊦ sort(a ' + x) = ins≤(a, sort(x));%a: ⊦ (x + y) + z = x + y + z; from specification list