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Proof step 7 from proof ole-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a0 < a, ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ ins≤(a, x0).first < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ [];%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