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Proof step 3 from proof ordered-append (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 4:
ordered≤(a + x1), ordered≤(x0), a + x1 ≠ [] ∧ x0 ≠ [] → ¬ x0.head < (a + x1).last, ∀ x0. ordered≤(x0) ∧ ordered≤(x1) ∧ (x1 ≠ [] ∧ x0 ≠ [] → ¬ x0.head < x1.last) → ordered≤(x1 + x0) ⊦ ordered≤((a + x1) + x0)

Rule argument

indvar: x

Additional information

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