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Proof step 13 from proof put-at (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1]
Proof step 15:
a0 ' + x0[m0, a] ≠ [] ∧ n ≠ 0, m0 < # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ (a0 ' + x0[m0, a]).rest[n - 1] = (a0 ' + x0)[n], m0 + 1 = n, x0 = []

Rule argument

spec: inst: name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1] subst: [n, x] → [n, a0 ' + x0[m0, a]] flags: rewrite, current

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%oz: ⊦ m + 1 ≠ 0; 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: n, a0 ' + x0[m0, a]