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Proof step 6 from proof firstn-first (previous step)

firstn(n, a ' + y) = a0 + y0 ⊦ a0 = a, firstn(n, a ' + y) = [], n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
firstn(n, a ' + y) = a0 ' + y0 ⊦ a0 = a, n = 0

Rule argument

Additional information

Used simplifier rules: firstN-empty: ⊦ firstn(n, x) = [] ↔ n = 0;% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ a + x = a ' + x; (s)