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Proof step 12 from proof firstN-lastN (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
1 + # x0 = n0 + n, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ firstn(n0 - 1, x0) + restn(n0, a ' + x0) = x0, n0 = 0

Rule argument

Additional information

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