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Proof step 3 from proof restn-rest (previous step)

∀ x. restn(n0, x).rest = restn(n0, x.rest) ⊦ restn(n0 +1, x).rest = restn(n0 +1, x.rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ x. restn(n0, x).rest = restn(n0, x.rest) ⊦ restn(n0, x.rest).rest = restn(n0, x.rest.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