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Proof step 1 from proof ins-yes (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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