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Proof step 9 from proof ole-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
ins≤(a, x0) = a1 ' + y, a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ordered≤(a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: empty: ⊦ ins≤(a, x) ≠ [];%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%c: ⊦ a + x = a ' + x; from specification list