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Proof step 7 from proof firstn-last (previous step)

∀ y, a. m ≤ # y → (a ' + firstn(m, y)).last = (a ' + y)[m], m +1 ≤ # y ⊦ (a ' + firstn(m +1, y)).last = (a ' + y)[m +1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m < # y, ∀ y, a. m ≤ # y → (a ' + firstn(m, y)).last = (a ' + y)[m] ⊦ (y.first ' + firstn(m, y.rest)).last = y[m], y = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%get-succ: ⊦ (a ' + x)[n + 1] = x[n];%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%l-05: ⊦ (b ' + a ' + x).last = (a ' + x).last; from specification list-last%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2