YES
proof of AProVE_07_kabasci05.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) QDPOrderProof [EQUIVALENT, 64 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 30 ms]
        (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (28) YES
    (29) QDP
        (30) QDPOrderProof [EQUIVALENT, 41 ms]
        (31) QDP
        (32) PisEmptyProof [EQUIVALENT, 0 ms]
        (33) YES


----------------------------------------

(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)
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)
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)
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(x_1, x_2)) =  	[[4A]] 	 +  	[[2A]] 	* 	x_1 	 +  	[[1A]] 	* 	x_2
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[-I]] 	 +  	[[5A]] 	* 	x_1 	 +  	[[2A]] 	* 	x_2
>>>

   <<<
 POL(s(x_1)) =  	[[2A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(TRANSFORM(x_1)) =  	[[4A]] 	 +  	[[2A]] 	* 	x_1
>>>

   <<<
 POL(transform(x_1)) =  	[[3A]] 	 +  	[[5A]] 	* 	x_1
>>>


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)
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(x_1, x_2)) = [2] + [1/2]x_1 + [1/2]x_2
   POL(TRANSFORM(x_1)) = [2] + x_1
   POL(cons(x_1, x_2)) = [4] + [2]x_1 + x_2
   POL(s(x_1)) = x_1
   POL(transform(x_1)) = [4]x_1
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)
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)
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)
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)
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)
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)
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)
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)
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)
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_2(x_1, x_2) ) = 2x_1 + x_2 + 2
POL( minus_2(x_1, x_2) ) = x_1
POL( max_2(x_1, x_2) ) = x_1 + x_2
POL( 0 ) = 0
POL( s_1(x_1) ) = 2x_1 + 1
POL( min_2(x_1, x_2) ) = x_1
POL( transform_1(x_1) ) = x_1 + 1
POL( cons_2(x_1, x_2) ) = 2x_1

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)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(33)
YES