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Proof step 5 from proof rest-sub (previous step)

# x = n0 + n ⊦ restn(n, x) = firstn(n0, restn(n, x))

User applied rule structural induction

And obtained the following new goals

Proof step 6:
# x = n0 + n1 +1, ∀ x, n0. # x = n0 + n1 → restn(n1, x) = firstn(n0, restn(n1, x)) ⊦ restn(n1 +1, x) = firstn(n0, restn(n1 +1, x))

Rule argument

indvar: n

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%first-all-01: ⊦ # x = n → firstn(n, x) = x;%nat freely generated by 0, +1;;% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic