(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(-(x1, x2)) = 2 + x1 + x2
POL(0) = 1
POL(double(x1)) = 1 + 2·x1
POL(half(x1)) = x1
POL(if(x1, x2, x3)) = 1 + 2·x1 + x2 + x3
POL(s(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
double(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
double(s(x)) → s(s(double(x)))
half(0) → 0
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(double(x1)) = x1
POL(half(x1)) = 1 + x1
POL(s(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
half(0) → 0
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
double(s(x)) → s(s(double(x)))
Q is empty.
(5) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
double(s(x)) → s(s(double(x)))
The set Q consists of the following terms:
double(s(x0))
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DOUBLE(s(x)) → DOUBLE(x)
The TRS R consists of the following rules:
double(s(x)) → s(s(double(x)))
The set Q consists of the following terms:
double(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(9) 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.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DOUBLE(s(x)) → DOUBLE(x)
R is empty.
The set Q consists of the following terms:
double(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(11) 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].
double(s(x0))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DOUBLE(s(x)) → DOUBLE(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- DOUBLE(s(x)) → DOUBLE(x)
The graph contains the following edges 1 > 1
(14) TRUE