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

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 Overlay + Local Confluence (⇔)
    ↳6 QTRS
    ↳7 DependencyPairsProof (⇔)
    ↳8 QDP
      ↳9 MNOCProof (⇔)
        ↳10 QDP
      ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 DependencyGraphProof (⇔)
        ↳14 AND
          ↳15 QDP
            ↳16 UsableRulesProof (⇔)
            ↳17 QDP
            ↳18 QReductionProof (⇔)
            ↳19 QDP
            ↳20 UsableRulesReductionPairsProof (⇔)
            ↳21 QDP
            ↳22 PisEmptyProof (⇔)
            ↳23 TRUE
          ↳24 QDP
            ↳25 UsableRulesProof (⇔)
            ↳26 QDP
            ↳27 QReductionProof (⇔)
            ↳28 QDP
            ↳29 UsableRulesReductionPairsProof (⇔)
            ↳30 QDP
            ↳31 PisEmptyProof (⇔)
            ↳32 TRUE

(0) Obligation:

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

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

d(a(x)) → a(b(c(d(x))))
c(b(x)) → b(a(d(c(x))))
c(a(x)) → x
d(b(x)) → x

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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

d(a(x)) → a(b(c(d(x))))
c(b(x)) → b(a(d(c(x))))
c(a(x)) → x
d(b(x)) → x

Q is empty.

(5) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(6) Obligation:

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

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

(7) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(8) Obligation:

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

A(d(x)) → B(a(x))
A(d(x)) → A(x)
B(c(x)) → A(b(x))
B(c(x)) → B(x)

The TRS R consists of the following rules:

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

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

(9) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(10) Obligation:

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

A(d(x)) → B(a(x))
A(d(x)) → A(x)
B(c(x)) → A(b(x))
B(c(x)) → B(x)

The TRS R consists of the following rules:

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A(d(x)) → B(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A(x1)) =
/0A\
\-I/
 +
/1A-I\
\-I-I/
·x1

POL(d(x1)) =
/0A\
\1A/
 +
/0A0A\
\0A1A/
·x1

POL(B(x1)) =
/0A\
\-I/
 +
/0A0A\
\-I-I/
·x1

POL(a(x1)) =
/0A\
\-I/
 +
/-I0A\
\-I0A/
·x1

POL(c(x1)) =
/1A\
\0A/
 +
/1A0A\
\0A0A/
·x1

POL(b(x1)) =
/0A\
\0A/
 +
/0A-I\
\0A-I/
·x1

The following usable rules [FROCOS05] were oriented:

b(d(x)) → x
a(c(x)) → x
b(c(x)) → c(d(a(b(x))))
a(d(x)) → d(c(b(a(x))))

(12) Obligation:

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

A(d(x)) → A(x)
B(c(x)) → A(b(x))
B(c(x)) → B(x)

The TRS R consists of the following rules:

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) Complex Obligation (AND)

(15) Obligation:

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

A(d(x)) → A(x)

The TRS R consists of the following rules:

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

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

(16) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

A(d(x)) → A(x)

R is empty.
The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

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

(18) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

(19) Obligation:

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

A(d(x)) → A(x)

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

(20) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

A(d(x)) → A(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(A(x1)) = 2·x1   
POL(d(x1)) = 2·x1   

(21) Obligation:

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

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

B(c(x)) → B(x)

The TRS R consists of the following rules:

a(d(x)) → d(c(b(a(x))))
b(c(x)) → c(d(a(b(x))))
a(c(x)) → x
b(d(x)) → x

The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

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

(25) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(26) Obligation:

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

B(c(x)) → B(x)

R is empty.
The set Q consists of the following terms:

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

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

(27) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

a(d(x0))
b(c(x0))
a(c(x0))
b(d(x0))

(28) Obligation:

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

B(c(x)) → B(x)

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

(29) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

B(c(x)) → B(x)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(B(x1)) = 2·x1   
POL(c(x1)) = 2·x1   

(30) Obligation:

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE