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Proof step 3 from proof firstn-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a0 ' + y0 = firstn(n, a0 ' + y0) + a ' + y, a0 ∈ firstn(n, a0 ' + y0) ∨ a0 = a ∨ a0 ∈ y ⊦ a0 ' + firstn(n, y0) = firstn(n, a0 ' + y0) + a '

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list