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Proof step 12 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';%empty: ⊦ ins≤(a, x) ≠ [];%a: ⊦ (x + y) + z = x + y + z; from specification list%c-03: ⊦ a ' = b ' + x ↔ a = b ∧ x = []; from specification list