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Proof step 12 from proof ordered-postfix (previous step)

x1 = a0 + y, ordered≤(x1), x ⊒ a ' + x1, ∀ x0. x0 ⊒ x1 → ordered≤(x0) ⊦ x1 = [], a0 < a, ordered≤(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
ordered≤(a0 ' + y), x ⊒ a ' + a0 ' + y, ∀ x0. x0 ⊒ a0 ' + y → ordered≤(x0) ⊦ a0 < a, ordered≤(x)

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% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)