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Proof step 8 from proof ordered-max (previous step)

max(x0) = x0.last, x0.last < a, #(a ' + x0) ≤ 1 ∨ ¬ (a ' + x0).tail.head < (a ' + x0).head ∧ ordered≤((a ' + x0).tail) ⊦ a = x0.last, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
max(x0) = x0.last, x0.last < a, ordered≤(x0) ⊦ a = x0.last, x0 = [], x0.head < a

Rule argument

Additional information

Used simplifier rules: lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem% ⊦ (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)