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Proof step 5 from proof restn-firstn (previous step)

∀ n0, x. m0 + n0 + m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)), m0 +1 + n0 + m0 +1 ≤ # x ⊦ restn(m0 +1, firstn(n0 + m0 +1, x)) = firstn(n0, restn(m0 +1, x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
2 + m0 + n0 + m0 ≤ # x, ∀ n0, x. m0 + n0 + m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, x.rest)) = firstn(n0, restn(m0, x.rest))

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%restN-rec: ⊦ restn(n + 1, x) = restn(n, x.rest);%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2