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Proof step 7 from proof sublist-part (previous step)

1 + # x0 = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ ((a ' + x0).head ' + firstn(m0 -1, (a ' + x0).tail)) + firstn(n0, restn(m0, a ' + x0)) = a ' + x0, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
1 + # x0 = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ firstn(m0 - 1, x0) + firstn(n0, restn(m0, a ' + x0)) = x0, m0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ x + y = x + z ↔ y = z; (s)% ⊦ (a ' + x).tail = x; (s)% ⊦ n -1 = n - 1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).head = a; (s)