YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/examples/collection/bincount.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 3 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 16 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 TRUE

(0) Obligation:

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

0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
P(0(x)) → P(1(0(0(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:

01(P(x)) → 11(P(x))
01(c(x)) → 11(0(x))
01(c(x)) → 01(x)
11(c(x)) → 01(x)
P1(0(x)) → P1(1(0(0(x))))
P1(0(x)) → 11(0(0(x)))
P1(0(x)) → 01(0(x))

The TRS R consists of the following rules:

0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
P(0(x)) → P(1(0(0(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 1 SCC with 3 less nodes.

(4) Obligation:

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

11(c(x)) → 01(x)
01(P(x)) → 11(P(x))
01(c(x)) → 11(0(x))
01(c(x)) → 01(x)

The TRS R consists of the following rules:

0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
P(0(x)) → P(1(0(0(x))))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


11(c(x)) → 01(x)
01(c(x)) → 01(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(01(x1)) = x1   
POL(1(x1)) = 1 + x1   
POL(11(x1)) = x1   
POL(P(x1)) = 0   
POL(c(x1)) = 1 + x1   

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

P(0(x)) → P(1(0(0(x))))
0(P(x)) → 1(P(x))
0(c(x)) → 1(0(x))
1(P(x)) → c(P(x))
1(c(x)) → c(0(x))

(6) Obligation:

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

01(P(x)) → 11(P(x))
01(c(x)) → 11(0(x))

The TRS R consists of the following rules:

0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
P(0(x)) → P(1(0(0(x))))

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 0 SCCs with 2 less nodes.

(8) TRUE