(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
*1(x, s(y)) → *1(x, y)
ODD(s(s(x))) → ODD(x)
HALF(s(s(x))) → HALF(x)
POW(x, y) → F(x, y, s(0))
F(x, s(y), z) → IF(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
F(x, s(y), z) → ODD(s(y))
F(x, s(y), z) → F(x, y, *(x, z))
F(x, s(y), z) → *1(x, z)
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
F(x, s(y), z) → *1(x, x)
F(x, s(y), z) → HALF(s(y))
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 6 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x))) → HALF(x)
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(x))) → HALF(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- HALF(s(s(x))) → HALF(x)
The graph contains the following edges 1 > 1
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ODD(s(s(x))) → ODD(x)
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ODD(s(s(x))) → ODD(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:
- ODD(s(s(x))) → ODD(x)
The graph contains the following edges 1 > 1
(14) YES
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, s(y)) → *1(x, y)
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(x, s(y)) → *1(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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(x, s(y)) → *1(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(19) YES
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
F(x, s(y), z) → F(x, y, *(x, z))
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
F(x, s(y), z) → F(x, y, *(x, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( F(x1, ..., x3) ) = max{0, 2x2 - 2} |
| POL( *(x1, x2) ) = 2x1 + 1 |
| POL( +(x1, x2) ) = max{0, 2x1 - 1} |
| POL( half(x1) ) = max{0, x1 - 2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
half(s(0)) → 0
half(s(s(x))) → s(half(x))
half(0) → 0
(22) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(24) YES
(25) 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
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(27) 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.
(28) 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
(29) YES