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

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

Heuristic simplifier applied rule simplifier

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, x.rest) = firstn(n0, restn(n1, x.rest))

Rule argument

Additional information

Used simplifier rules: restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2