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Proof step 5 from proof put-length (previous step)

n ≤ # x0, ∀ n, a. n < # x0 → # x0[n, a] = # x0 ∨ x0 = [] ⊦ # (a0 ' + x0)[n, a] = 1 + # x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a]
Proof step 7:
n ≤ # x0, ∀ n, a. n < # x0 → # x0[n, a] = # x0 ∨ x0 = [] ⊦ #((a0 ' + x0).head ' + (a0 ' + x0).tail[n -1, a]) = 1 + # x0, a0 ' + x0 = [], n = 0

Rule argument

spec: <toplevel> name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a] subst: [a, n, x] → [a, n, a0 ' + x0] flags: <all paths>

Additional information

Used simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s) Used terms: a, n, a0 ' + x0