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Proof step 23 from proof max-ins-max (previous step)

max(ins≤(a, x0)) < a, a0 < a, max(ins≤(a, x0)) < a0, ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ a0 = a, a < max(a0 ' + x0), #(a ' + x0) ≤ 1, ¬ (a ' + x0).tail.head < (a ' + x0).head ∧ ordered≤((a ' + x0).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
max(ins≤(a, x0)) < a0, a0 < a, max(ins≤(a, x0)) < a, ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ a0 = a, x0 = [], a < (max(x0) < a0 ⊃ a0;max(x0)), ¬ x0.head < a ∧ ordered≤(x0)

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x));% ⊦ (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)