(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(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, y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
GCD(s(x), s(y)) → MINUS(max(x, y), min(x, transform(y)))
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, transform(y))
GCD(s(x), s(y)) → TRANSFORM(y)
GCD(s(x), s(y)) → MIN(x, y)
TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
TRANSFORM(cons(x, y)) → CONS(x, x)
TRANSFORM(s(x)) → TRANSFORM(x)
CONS(x, cons(y, s(z))) → CONS(y, x)
CONS(cons(x, z), s(y)) → TRANSFORM(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(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 5 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(x, cons(y, s(z))) → CONS(y, x)
CONS(cons(x, z), s(y)) → TRANSFORM(x)
TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
TRANSFORM(cons(x, y)) → CONS(x, x)
TRANSFORM(s(x)) → TRANSFORM(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
CONS(x, cons(y, s(z))) → CONS(y, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(CONS(x1, x2)) = | 4A | + | 2A | · | x1 | + | 1A | · | x2 |
| POL(cons(x1, x2)) = | -I | + | 5A | · | x1 | + | 2A | · | x2 |
| POL(TRANSFORM(x1)) = | 4A | + | 2A | · | x1 |
| POL(transform(x1)) = | 3A | + | 5A | · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
cons(x, y) → y
transform(cons(x, y)) → cons(cons(x, x), x)
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
transform(x) → s(s(x))
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(cons(x, z), s(y)) → TRANSFORM(x)
TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
TRANSFORM(cons(x, y)) → CONS(x, x)
TRANSFORM(s(x)) → TRANSFORM(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
CONS(cons(x, z), s(y)) → TRANSFORM(x)
TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
TRANSFORM(cons(x, y)) → CONS(x, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:
POL(CONS(x1, x2)) = [2] + [1/2]x1 + [1/2]x2
POL(TRANSFORM(x1)) = [2] + x1
POL(cons(x1, x2)) = [4] + [2]x1 + x2
POL(s(x1)) = x1
POL(transform(x1)) = [4]x1
The value of delta used in the strict ordering is 2.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
cons(x, y) → y
transform(cons(x, y)) → cons(cons(x, x), x)
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
transform(x) → s(s(x))
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TRANSFORM(s(x)) → TRANSFORM(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TRANSFORM(s(x)) → TRANSFORM(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:
- TRANSFORM(s(x)) → TRANSFORM(x)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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, y)
The graph contains the following edges 1 > 1, 2 > 2
(18) YES
(19) 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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) 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.
(21) 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.
(22) 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
(23) YES
(24) 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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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.
(26) 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.
(27) 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
(28) YES
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, transform(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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) 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, transform(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 |
| POL( transform(x1) ) = x1 + 1 |
| POL( cons(x1, x2) ) = 2x1 |
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, y))
(31) 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, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(32) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(33) YES