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Proof step 7 from proof ordered-max (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
max(x0) = x0.last, ordered≤(x0), x0.last < a ⊦ x0 = [], a = x0.last, x0.first < a

Rule argument

Additional information

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