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Proof step 10 from proof firstn-last-rec (previous step)

m ≤ # y0, ∀ x0. m < # x0 → x0.head ' + firstn(m, x0.tail) = firstn(m, x0) + x0[m] ' ⊦ a ' + firstn(m, y0) = firstn(m, a ' + y0) + (a ' + y0)[m] '

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a ' + y0

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (a ' + x).head = a; (s)% ⊦ (a ' + x).tail = x; (s) Used terms: a ' + y0