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Proof step 6 from proof firstn-firstn (previous step)

x = a + y, m0 < n, m0 < # x, 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) = a ' + firstn(m0, y), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 < n, m0 ≤ # y, n ≤ 1 + # y, ∀ n, x. n ≤ # x ∧ m0 ≤ n ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(n, a ' + y).head ' + firstn(m0, firstn(n, a ' + y).tail) = a ' + firstn(m0, y)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)