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Proof step 14 from proof eq (previous step)

a0 ' + ins≤(a, x0) = y.head + ins≤(a, y.tail), a0 < a, y.head < a, a0 ∈ y.head + ins≤(a, y.tail), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
ins≤(a, x0) = ins≤(a, y.tail), y.head < a, ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ y.head ' + x0 = y, y = []

Rule argument

Additional information

Used simplifier rules: lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)% ⊦ a + x = a ' + x; (s)