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Proof step 15 from proof ordered-prefix (previous step)

x1 ⊑ y, (∃ x0. ordered≤(x0) ∧ x1 ⊑ x0) → ordered≤(x1), y = [] ∨ ¬ y.head < a ∧ ordered≤(y) ⊦ #(a ' + x1) ≤ 1, ¬ (a ' + x1).tail.head < (a ' + x1).head ∧ ordered≤((a ' + x1).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
ordered≤(y), x1 ⊑ y, (∃ x0. ordered≤(x0) ∧ x1 ⊑ x0) → ordered≤(x1) ⊦ x1 = [], y = [], y.head < a, ¬ x1.head < a ∧ ordered≤(x1)

Rule argument

Additional information

Used simplifier rules: ⊦ x ⊑ [] ↔ x = []; (s)% ⊦ (a ' + x).head = a; (s)% ⊦ (a ' + x).tail = x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)