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Proof step 14 from proof put-sw-in (previous step)

m0 < # x0, m ≤ # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ a ∈ (a0 ' + x0)[m, x0[m0]][m0 + 1, (a0 ' + x0)[m]] ↔ a ∈ a0 ' + x0, x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1]
Proof step 16:
m0 < # x0, m ≤ # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ a0 ' + x0 = [], a ∈ (a0 ' + x0)[m, x0[m0]][m0 + 1, (a0 ' + x0).tail[m - 1]] ↔ a ∈ a0 ' + x0, m = 0, x0 = []

Rule argument

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

Additional information

Used simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;%get-zero-01: ⊦ (a ' + x)[0] = a;%put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];%put-in: ⊦ a ≠ x[m] ∧ m < # x → (a ∈ x[m, b] ↔ a = b ∨ a ∈ x);%get-in: ⊦ m < # x → (x[m] ∈ x ↔ true);% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s)% ⊦ a + x = a ' + x; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, a0 ' + x0