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Proof step 12 from proof ole-01 (previous step)

a0 ' + x0 ≠ [] ∧ (a0 ' + x0).rest ≠ [], ins≤(a, x0) = a1 ' + y, a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ (a0 ' + x0).rest.first < (a0 ' + x0).first ∧ ordered≤((a0 ' + x0).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
ins≤(a, x0) = a1 ' + y, a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ x0.first < a0 ∧ ordered≤(x0), x0 = []

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list