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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 Instantiation (⇔)
↳4 QDP
↳5 Narrowing (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 Instantiation (⇔)
↳10 QDP
↳11 DependencyGraphProof (⇔)
↳12 QDP
↳13 QDPOrderProof (⇔)
↳14 QDP
↳15 Narrowing (⇔)
↳16 QDP
↳17 DependencyGraphProof (⇔)
↳18 QDP
↳19 UsableRulesProof (⇔)
↳20 QDP
↳21 Instantiation (⇔)
↳22 QDP
↳23 Instantiation (⇔)
↳24 QDP
↳25 DependencyGraphProof (⇔)
↳26 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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:

B(x, y) → A(c(y), a(0, x))
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(3) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule B(x, y) → A(c(y), a(0, x)) we obtained the following new rules [LPAR04]:

B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y_0) → A(c(y_0), a(0, 0))

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y_0) → A(c(y_0), a(0, 0))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(5) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule B(0, y_0) → A(c(y_0), a(0, 0)) at position [1] we obtained the following new rules [LPAR04]:

B(0, y0) → A(c(y0), 0)

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))
B(0, y0) → A(c(y0), 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(x, y) → A(0, x)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(9) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule B(x, y) → A(0, x) we obtained the following new rules [LPAR04]:

B(y_3, 0) → A(0, y_3)
B(0, y_0) → A(0, 0)

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
B(0, y_0) → A(0, 0)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(y_3, 0) → A(c(0), a(0, y_3))
A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
A(y, c(b(a(0, x), 0))) → B(0, y)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A(y, c(b(a(0, x), 0))) → B(a(c(b(0, y)), x), 0)
A(y, c(b(a(0, x), 0))) → A(c(b(0, y)), x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(B(x1, x2)) =
/1\
\1/
 +
/10\
\10/
·x1 +
/00\
\10/
·x2

POL(0) =
/1\
\0/

POL(A(x1, x2)) =
/0\
\0/
 +
/10\
\11/
·x1 +
/10\
\10/
·x2

POL(c(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(a(x1, x2)) =
/0\
\0/
 +
/10\
\11/
·x1 +
/10\
\10/
·x2

POL(b(x1, x2)) =
/1\
\0/
 +
/10\
\01/
·x1 +
/01\
\10/
·x2

The following usable rules [FROCOS05] were oriented:

a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)
b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(y_3, 0) → A(c(0), a(0, y_3))
B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → B(0, y)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(15) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule B(y_3, 0) → A(c(0), a(0, y_3)) at position [1] we obtained the following new rules [LPAR04]:

B(x1, 0) → A(c(0), 0)
B(c(b(a(0, x1), 0)), 0) → A(c(0), b(a(c(b(0, 0)), x1), 0))

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(y_3, 0) → A(0, y_3)
A(y, c(b(a(0, x), 0))) → B(0, y)
B(x1, 0) → A(c(0), 0)
B(c(b(a(0, x1), 0)), 0) → A(c(0), b(a(c(b(0, 0)), x1), 0))

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(0, y_3)

The TRS R consists of the following rules:

b(x, y) → c(a(c(y), a(0, x)))
a(y, x) → y
a(y, c(b(a(0, x), 0))) → b(a(c(b(0, y)), x), 0)

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

(19) 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.

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(y, c(b(a(0, x), 0))) → B(0, y)
B(y_3, 0) → A(0, y_3)

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

(21) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule A(y, c(b(a(0, x), 0))) → B(0, y) we obtained the following new rules [LPAR04]:

A(0, c(b(a(0, x1), 0))) → B(0, 0)

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

B(y_3, 0) → A(0, y_3)
A(0, c(b(a(0, x1), 0))) → B(0, 0)

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

(23) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule B(y_3, 0) → A(0, y_3) we obtained the following new rules [LPAR04]:

B(0, 0) → A(0, 0)

(24) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(0, c(b(a(0, x1), 0))) → B(0, 0)
B(0, 0) → A(0, 0)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(26) TRUE