[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 14 from proof put-at (previous step)

m0 < # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ m0 + 1 = n, (a0 ' + x0[m0, a])[n] = (a0 ' + x0)[n], 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, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ m0 + 1 = n, a0 ' + x0[m0, a] = [], (a0 ' + x0[m0, a]).tail[n - 1] = (a0 ' + x0)[n], n = 0, x0 = []

Rule argument

spec: <toplevel> name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.tail[n - 1] subst: [n, x] → [n, a0 ' + x0[m0, a]] flags: <all paths>

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, a0 ' + x0[m0, a]