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

x1 ⊑ y, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1), y = [] ∨ ¬ y.first < a ∧ ordered≤(y) ⊦ #(a ' + x1) ≤ 1 ∨ ¬ (a ' + x1).rest.first < (a ' + x1).first ∧ ordered≤((a ' + x1).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: prefix-empty: ⊦ x ⊑ [] ↔ x = []; from specification list-last%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