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Proof step 2 from proof sublist-last (previous step)

m0 + n < # x ⊦ (restn(m0, x).first ' + firstn(n, restn(m0, x).rest)).last = x[m0 + n], x = []

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [], m0 + n0 +1 < # x ⊦ (restn(m0, x).first ' + firstn(n0 +1, restn(m0, x).rest)).last = x[m0 + n0 +1], x = []

Rule argument

indvar: n

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%restn-first-get: ⊦ n < # x → restn(n, x).first = x[n];%nat freely generated by 0, +1;;%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%l: ⊦ a '.last = a; from specification list-last%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2