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Proof step 5 from proof firstn-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
firstn(n, a ' + y) = a0 ' + y0 ⊦ n = 0, a0 = a

Rule argument

Additional information

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