(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
The set Q consists of the following terms:
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
F(s(x)) → -1(s(x), g(f(x)))
F(s(x)) → G(f(x))
F(s(x)) → F(x)
G(s(x)) → -1(s(x), f(g(x)))
G(s(x)) → F(g(x))
G(s(x)) → G(x)
The TRS R consists of the following rules:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
The set Q consists of the following terms:
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
The TRS R consists of the following rules:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
The set Q consists of the following terms:
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
R is empty.
The set Q consists of the following terms:
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- -1(s(x), s(y)) → -1(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → G(f(x))
G(s(x)) → F(g(x))
F(s(x)) → F(x)
G(s(x)) → G(x)
The TRS R consists of the following rules:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
The set Q consists of the following terms:
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(s(x)) → G(f(x))
G(s(x)) → F(g(x))
F(s(x)) → F(x)
G(s(x)) → G(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1) =
x1
s(
x1) =
s(
x1)
G(
x1) =
G(
x1)
f(
x1) =
x1
g(
x1) =
g(
x1)
0 =
0
-(
x1,
x2) =
x1
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s_1=3
0=1
G_1=2
g_1=4
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
-(0, s(y)) → 0
(16) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))
The set Q consists of the following terms:
-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) YES