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

a0 ' + y0 = firstn(n0 +1, a0 ' + y0) + a ' + y, ∀ a0, y0, a, y. a0 ' + y0 = firstn(n0, a0 ' + y0) + a ' + y ∧ (a0 ∈ firstn(n0, a0 ' + y0) ∨ a0 = a ∨ a0 ∈ y) → a0 ' + firstn(n0, y0) = firstn(n0, a0 ' + y0) + a ', a0 ∈ firstn(n0 +1, a0 ' + y0) ∨ a0 = a ∨ a0 ∈ y ⊦ a0 ' + firstn(n0 +1, y0) = firstn(n0 +1, a0 ' + y0) + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
y0 = firstn(n0, y0) + a ' + y, ∀ a0, y0, a, y. a0 ' + y0 = firstn(n0, a0 ' + y0) + a ' + y ∧ (a0 ∈ firstn(n0, a0 ' + y0) ∨ a0 = a ∨ a0 ∈ y) → a0 ' + firstn(n0, y0) = firstn(n0, a0 ' + y0) + a ' ⊦ y0.head ' + firstn(n0, y0.tail) = firstn(n0, y0) + a '

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ x + y = x + z ↔ y = z; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).head = a; (s)% ⊦ m +1 = m + 1; (s)