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Proof step 4 from proof sublist-last (previous step)

m0 + n0 +1 < # x, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).head ' + firstn(n0, restn(m0, x).tail)).last = x[m0 + n0] ∨ x = [] ⊦ (restn(m0, x).head ' + firstn(n0 +1, restn(m0, x).tail)).last = x[m0 + n0 +1], x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 + n0 + 1 < # x, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).head ' + firstn(n0, restn(m0, x).tail)).last = x[m0 + n0] ∨ x = [] ⊦ (restn(m0, x).tail.head ' + firstn(n0, restn(m0, x).tail.tail)).last = x.tail[m0 + n0], x = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);%get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.tail[n];% ⊦ (b ' + a ' + x).last = (a ' + x).last; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m +1 = m + 1; (s)