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Proof step 2 from proof ins-less-first (previous step)

ordered≤(x) ⊦ ins≤(a, x) = x.head + ins≤(a, x.tail), x = [], a < x.head

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
ordered≤(x) ⊦ ins≤(a, x) = x.head ' + ins≤(a, x.tail), x = [], a < x.head

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)