(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
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:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
The set Q consists of the following terms:
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
(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:
F(s(x)) → F(x)
G(s(x), s(y)) → IF(f(x), s(x), s(y))
G(s(x), s(y)) → F(x)
G(x, c(y)) → G(x, g(s(c(y)), y))
G(x, c(y)) → G(s(c(y)), y)
The TRS R consists of the following rules:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
The set Q consists of the following terms:
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
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:
F(s(x)) → F(x)
The TRS R consists of the following rules:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
The set Q consists of the following terms:
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
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:
F(s(x)) → F(x)
R is empty.
The set Q consists of the following terms:
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
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].
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x)) → F(x)
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:
- F(s(x)) → F(x)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(x, c(y)) → G(s(c(y)), y)
G(x, c(y)) → G(x, g(s(c(y)), y))
The TRS R consists of the following rules:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
The set Q consists of the following terms:
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
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.
G(x, c(y)) → G(s(c(y)), y)
G(x, c(y)) → G(x, g(s(c(y)), y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(
x1,
x2) =
x2
c(
x1) =
c(
x1)
g(
x1,
x2) =
g
if(
x1,
x2,
x3) =
if(
x2,
x3)
s(
x1) =
s
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s=1
c_1=4
g=4
if_2=1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
if(true, x, y) → x
if(false, x, y) → y
(16) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))
The set Q consists of the following terms:
f(0)
f(1)
f(s(x0))
if(true, x0, x1)
if(false, x0, x1)
g(s(x0), s(x1))
g(x0, c(x1))
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