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Proof step 4 from proof rest-len (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → # restn(n1, x) = n0 ⊦ # restn(n1, x.tail) = n0

Rule argument

Additional information

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