YES
proof of Secret_06_TRS_gen-25.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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) QDP
(5) QDPOrderProof [EQUIVALENT, 515 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 468 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(a, f(b(b(z, y), a))) -> z
   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))

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:

   C(c(z, x, a), a, y) -> F(f(c(y, a, f(c(z, y, x)))))
   C(c(z, x, a), a, y) -> F(c(y, a, f(c(z, y, x))))
   C(c(z, x, a), a, y) -> C(y, a, f(c(z, y, x)))
   C(c(z, x, a), a, y) -> F(c(z, y, x))
   C(c(z, x, a), a, y) -> C(z, y, x)
   F(f(c(a, y, z))) -> B(y, b(z, z))
   F(f(c(a, y, z))) -> B(z, z)

The TRS R consists of the following rules:

   b(a, f(b(b(z, y), a))) -> z
   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))

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 1 SCC with 5 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(c(z, x, a), a, y) -> C(z, y, x)
   C(c(z, x, a), a, y) -> C(y, a, f(c(z, y, x)))

The TRS R consists of the following rules:

   b(a, f(b(b(z, y), a))) -> z
   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   C(c(z, x, a), a, y) -> C(z, y, x)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(C(x_1, x_2, x_3)) =  	[[1A]] 	 +  	[[-I, 3A, 1A]] 	* 	x_1 	 +  	[[-I, 1A, 2A]] 	* 	x_2 	 +  	[[-I, 3A, 3A]] 	* 	x_3
>>>

   <<<
 POL(c(x_1, x_2, x_3)) =  	[[2A], [2A], [0A]] 	 +  	[[0A, -I, 3A], [-I, 2A, 1A], [-I, 0A, 0A]] 	* 	x_1 	 +  	[[0A, -I, -I], [-I, 1A, 2A], [-I, 0A, 0A]] 	* 	x_2 	 +  	[[0A, 3A, 3A], [-I, 1A, 0A], [-I, 0A, 0A]] 	* 	x_3
>>>

   <<<
 POL(a) =  	[[1A], [2A], [-I]]
>>>

   <<<
 POL(f(x_1)) =  	[[0A], [0A], [-I]] 	 +  	[[0A, 1A, 1A], [-I, -I, 0A], [-I, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(b(x_1, x_2)) =  	[[-I], [2A], [-I]] 	 +  	[[0A, -I, 0A], [-I, 0A, 0A], [-I, 0A, -I]] 	* 	x_1 	 +  	[[0A, -I, 0A], [-I, 0A, -I], [-I, 0A, 0A]] 	* 	x_2
>>>


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))
   b(a, f(b(b(z, y), a))) -> z


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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(c(z, x, a), a, y) -> C(y, a, f(c(z, y, x)))

The TRS R consists of the following rules:

   b(a, f(b(b(z, y), a))) -> z
   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(7) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   C(c(z, x, a), a, y) -> C(y, a, f(c(z, y, x)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(C(x_1, x_2, x_3)) =  	[[-I]] 	 +  	[[-I, -I, 1A]] 	* 	x_1 	 +  	[[0A, -I, -I]] 	* 	x_2 	 +  	[[-I, -I, 3A]] 	* 	x_3
>>>

   <<<
 POL(c(x_1, x_2, x_3)) =  	[[0A], [-I], [0A]] 	 +  	[[-I, -I, -I], [-I, 0A, 0A], [-I, -I, -I]] 	* 	x_1 	 +  	[[-I, -I, -I], [-I, 0A, 0A], [0A, 0A, -I]] 	* 	x_2 	 +  	[[-I, -I, -I], [0A, 0A, 0A], [-I, -I, 0A]] 	* 	x_3
>>>

   <<<
 POL(a) =  	[[0A], [0A], [3A]]
>>>

   <<<
 POL(f(x_1)) =  	[[0A], [0A], [-I]] 	 +  	[[0A, -I, 0A], [-I, 0A, -I], [0A, -I, -I]] 	* 	x_1
>>>

   <<<
 POL(b(x_1, x_2)) =  	[[-I], [0A], [0A]] 	 +  	[[0A, 0A, -I], [-I, -I, 0A], [-I, -I, -I]] 	* 	x_1 	 +  	[[-I, -I, 0A], [0A, -I, -I], [-I, -I, 0A]] 	* 	x_2
>>>


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))
   b(a, f(b(b(z, y), a))) -> z


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

(8)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   b(a, f(b(b(z, y), a))) -> z
   c(c(z, x, a), a, y) -> f(f(c(y, a, f(c(z, y, x)))))
   f(f(c(a, y, z))) -> b(y, b(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(10)
YES