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Proof step 4 from proof put-length (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.first ' + x.rest[n -1, a]
Proof step 6:
n ≠ 0 ∧ a0 ' + x0 ≠ [], n ≤ # x0, ∀ n, a. n < # x0 → # x0[n, a] = # x0 ∨ x0 = [] ⊦ #((a0 ' + x0).first ' + (a0 ' + x0).rest[n -1, a]) = # x0 + 1

Rule argument

spec: list-del inst: name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.first ' + x.rest[n -1, a] subst: [a, n, x] → [a, n, a0 ' + x0] flags: rewrite, current

Additional information

Used simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a, n, a0 ' + x0