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Proof step 14 from proof ole-01 (previous step)

ins≤(a, x0) = a1 ' + y, x0 = a2 + y0, a0 < a, a1 ∈ ins≤(a, x0), ∀ a. ordered≤(ins≤(a, x0)) ↔ ordered≤(x0) ⊦ ¬ a1 < a0 ∧ ordered≤(ins≤(a, x0)) ↔ ¬ a2 < a0 ∧ ordered≤(x0), x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
ins≤(a, a2 ' + y0) = a1 ' + y, a0 < a, ∀ a3. ordered≤(ins≤(a3, a2 ' + y0)) ↔ ordered≤(a2 ' + y0) ⊦ ¬ a1 < a0 ∧ ordered≤(a1 ' + y) ↔ ¬ a2 < a0 ∧ ordered≤(a2 ' + y0)

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)