(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
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:
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
MINUS(s(x), s(y)) → MINUS(x, any(y))
MINUS(s(x), s(y)) → ANY(y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → MINUS(max(x, y), min(x, y))
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
ANY(s(x)) → ANY(x)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
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 4 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ANY(s(x)) → ANY(x)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
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:
ANY(s(x)) → ANY(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:
- ANY(s(x)) → ANY(x)
The graph contains the following edges 1 > 1
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, any(y))
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
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:
MINUS(s(x), s(y)) → MINUS(x, any(y))
The TRS R consists of the following rules:
any(s(x)) → s(s(any(x)))
any(x) → x
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:
- MINUS(s(x), s(y)) → MINUS(x, any(y))
The graph contains the following edges 1 > 1
(14) YES
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
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:
MAX(s(x), s(y)) → MAX(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:
- MAX(s(x), s(y)) → MAX(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:
MIN(s(x), s(y)) → MIN(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) 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:
- MIN(s(x), s(y)) → MIN(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(24) YES
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( GCD(x1, x2) ) = 2x1 + x2 + 2 |
| POL( minus(x1, x2) ) = x1 |
| POL( max(x1, x2) ) = x1 + x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
(27) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(29) YES