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Proof step 3 from proof ins-max (previous step)

ordered≤(x0) ⊦ ins≤(a, x0) = x0 + a ', a < max(x0)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
ordered≤(a0 + x), ∀ a. ordered≤(x) → ins≤(a, x) = x + a ' ∨ a < max(x) ⊦ ins≤(a, a0 + x) = (a0 + x) + a ', a < max(a0 + x)

Rule argument

term: x0

Additional information

Used simplifier rules: list freely generated by [], +;;% ⊦ ins≤(a, []) = a '; (s)% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' = b ' ↔ a = b; (s)