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Proof step 1 from proof ins-less-first (conclusion)

⊦ ordered≤(x) ∧ x ≠ [] ∧ ¬ a < x.first → ins≤(a, x) = x.first + ins≤(a, x.rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
ordered≤(x) ⊦ x = [], ins≤(a, x) = x.first ' + ins≤(a, x.rest), a < x.first

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list