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Proof step 3 from proof firstn-last (previous step)

m = m0 + 1, m ≤ # x ⊦ m = 0, firstn(m, x).last = x[m0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 < # x ⊦ (x.first ' + firstn(m0, x.rest)).last = x[m0], x = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%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%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2