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Proof step 13 from proof firstN-lastN (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# x0 = m + n, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ firstn(m, x0) + restn(m, x0) = x0

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2