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Proof step 21 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
a ' + a0 ' + x0 = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y, y = [], a0 < a, y.first < a

Rule argument

Additional information

Used simplifier rules: ins-e: ⊦ ins≤(a, []) = a ';%a: ⊦ (x + y) + z = x + y + z; from specification list%e: ⊦ a ' ≠ []; from specification list%one-01: ⊦ y ≠ [] → (a ' = x + y ↔ x = [] ∧ a ' = y); from specification list