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Proof step 8 from proof firstN-lastN (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# x0 + 1 = n0 + n, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ firstn(n0, a ' + x0) + restn(n0, a ' + x0) = a ' + x0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list