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

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

(0) Obligation:

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

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 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:

TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))
TIMES(x, plus(y, 1)) → TIMES(1, 0)

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

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:

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 1   
POL(TIMES(x1, x2)) = x2   
POL(plus(x1, x2)) = x1 + x2   
POL(times(x1, x2)) = 0   

The following usable rules [FROCOS05] were oriented:

times(x, 0) → 0
plus(x, 0) → x

(6) Obligation:

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

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE