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Proof step 5 from proof ordered-append (previous step)

x0 = a0 + y ∧ x0 ≠ [], ordered≤(a ' + x1), ordered≤(x0), x0 ≠ [] → ¬ a0 < (a ' + x1).last, ∀ x. ordered≤(x) ∧ (x1 ≠ [] ∧ x ≠ [] → ¬ x.first < x1.last) → ordered≤(x1 + x) ⊦ ordered≤((a ' + x1) + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ordered≤(a0 ' + y), ordered≤(a ' + x1), ∀ x. ordered≤(x) ∧ (x1 ≠ [] ∧ x ≠ [] → ¬ x.first < x1.last) → ordered≤(x1 + x) ⊦ a0 < (a ' + x1).last, ordered≤(a ' + x1 + a0 ' + y)

Rule argument

Additional information

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