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Proof step 16 from proof ordered-prefix (previous step)

y = a0 + y0, x1 ⊑ y, ordered≤(y), (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ x1 = [], y = [], a0 < a, ¬ x1.first < a ∧ ordered≤(x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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