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Proof step 16 from proof eq (previous step)

ins≤(a, x0) = ins≤(a, y0), y = a0 + y0, a0 < a, ∀ 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 17:
ins≤(a, x0) = ins≤(a, y0), a0 < a, ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ x0 = y0

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)