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Proof step 22 from proof max-ins-max (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
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.first < a ∧ ordered≤(x0)

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x));%c-08: ⊦ (a ' + x).first = a; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list