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Proof step 2 from proof ordered-append (previous step)

ordered≤(x), ordered≤(x0), x ≠ [] ∧ x0 ≠ [] → ¬ x0.first < x.last ⊦ ordered≤(x + x0)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ x0. ordered≤(x1) ∧ ordered≤(x0) ∧ (x1 ≠ [] ∧ x0 ≠ [] → ¬ x0.first < x1.last) → ordered≤(x1 + x0), ordered≤(a + x1), ordered≤(x0), a + x1 ≠ [] ∧ x0 ≠ [] → ¬ x0.first < (a + x1).last ⊦ ordered≤((a + x1) + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: le-e: ⊦ ordered≤([]);%list freely generated by [], +;;%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list