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Proof step 9 from proof sublist-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m0 + n0 < # y, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).head ' + firstn(n0, restn(m0, x).tail)).last = x[m0 + n0] ∨ x = [] ⊦ (restn(m0, y).head ' + firstn(n0, restn(m0, y).tail)).last = y[m0 + n0], y = []

Rule argument

Additional information

Used simplifier rules: restn-apprec: ⊦ restn(m, restn(n, x)) = restn(m + n, x);%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%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)