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Proof step 8 from proof sublist-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m0 + n0 < # y, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [] ⊦ y = [], (restn(m0, y).first ' + firstn(n0, restn(m0, y).rest)).last = y[m0 + n0]

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