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Proof step 9 from proof sublist-part (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# x0 = n0 + m, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ firstn(m, x0) + firstn(n0, restn(m, x0)) = x0

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.tail);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)