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Proof step 14 from proof ordered-append (previous step)

ordered≤(x1 + a0 ' + y), ordered≤(a ' + x1), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], a0 < x1.last, #(a ' + x1 + a0 ' + y) ≤ 1, ¬ (a ' + x1 + a0 ' + y).tail.head < (a ' + x1 + a0 ' + y).head ∧ ordered≤((a ' + x1 + a0 ' + y).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
x1.head < a, ordered≤(a ' + x1), ordered≤(x1 + a0 ' + y), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.head < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], a0 < x1.last

Rule argument

Additional information

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