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

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: app-01: ⊦ ordered≤(x + y) → ordered≤(y);%cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list