(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Combined order from the following AFS and order.
app(
x1,
x2) =
app(
x1,
x2)
nil =
nil
cons(
x1,
x2) =
cons(
x1,
x2)
sum(
x1) =
x1
a(
x1,
x2,
x3) =
a(
x1,
x2,
x3)
h =
h
s(
x1) =
s(
x1)
Recursive path order with status [RPO].
Quasi-Precedence:
app2 > [nil, cons2, h] > a3 > s1
Status:
app2: multiset
nil: multiset
cons2: [2,1]
a3: [1,2,3]
h: multiset
s1: multiset
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
sum(cons(x, nil)) → cons(x, nil)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Knuth-Bendix order [KBO] with precedence:
sum1 > nil > cons2
and weight map:
nil=1
sum_1=0
cons_2=0
The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
sum(cons(x, nil)) → cons(x, nil)
(4) Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.
(5) RisEmptyProof (EQUIVALENT transformation)
The TRS R is empty. Hence, termination is trivially proven.
(6) YES