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Proof step 11 from proof ordered-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
ordered≤(a ' + x1), ordered≤(x1 + a0 ' + y), ∀ 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: app-01: ⊦ ordered≤(x + y) → ordered≤(y);% ⊦ (a ' + x).head = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)