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Proof step 4 from proof firstn-firstn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 < # x, m0 < n, n ≤ # x, ∀ n, x. n ≤ # x ∧ m0 ≤ n ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(n, x).head ' + firstn(m0, firstn(n, x).tail) = x.head ' + firstn(m0, x.tail), x = []

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ m < n → n ≠ 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m +1 = m + 1; (s)