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

∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x), m0 +1 ≤ n, n ≤ # x, m0 +1 ≤ # x ⊦ firstn(m0 +1, firstn(n, x)) = firstn(m0 +1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 < # x, n ≤ # x, m0 < n, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ x = [], n = 0, firstn(m0, firstn(n, x).rest) = firstn(m0, x.rest)

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%firstn-first: ⊦ n ≠ 0 ∧ x ≠ [] → firstn(n, x).first = x.first;%a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%l: ⊦ # x = 0 ↔ x = []; from specification list-data%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%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