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Proof step 9 from proof ordered-append (previous step)

ordered≤(a ' + x1), ordered≤(a0 ' + y), ∀ x. ordered≤(x) ∧ (x1 ≠ [] ∧ x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], a0 < (a ' + x1).last, ordered≤(a ' + x1 + a0 ' + y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ordered≤(a0 ' + y), ordered≤(a ' + x1), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], a0 < x1.last, ordered≤(a ' + x1 + a0 ' + y)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x ≠ [] → (a ' + x).last = x.last; (s)