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

∀ n0, y. n0 ≤ # y → restn(n0 + # x0, x0 + y) = restn(n0, y), n0 ≤ # y ⊦ restn(n0 + #(a + x0), (a + x0) + y) = restn(n0, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list