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 DependencyGraphProof (⇔)
↳8 QDP
↳9 QDPOrderProof (⇔)
↳10 QDP
↳11 PisEmptyProof (⇔)
↳12 TRUE

(0) Obligation:

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

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

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:

ACK_IN(s(m), 0) → U11(ack_in(m, s(0)))
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)
ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
U21(ack_out(n), m) → U22(ack_in(m, n))
U21(ack_out(n), m) → ACK_IN(m, n)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

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 2 less nodes.

(4) Obligation:

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

ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)
U21(ack_out(n), m) → ACK_IN(m, n)
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

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.


U21(ack_out(n), m) → ACK_IN(m, n)
ACK_IN(s(m), 0) → ACK_IN(m, s(0))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(ACK_IN(x1, x2)) = x1   
POL(U21(x1, x2)) = 1 + x2   
POL(ack_in(x1, x2)) = 1 + x1 + x2   
POL(ack_out(x1)) = 1   
POL(s(x1)) = 1 + x1   
POL(u11(x1)) = 0   
POL(u21(x1, x2)) = 0   
POL(u22(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

ACK_IN(s(m), s(n)) → U21(ack_in(s(m), n), m)
ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

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:

ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)

The TRS R consists of the following rules:

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACK_IN(s(m), s(n)) → ACK_IN(s(m), n)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACK_IN(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

ack_in(0, n) → ack_out(s(n))
ack_in(s(m), 0) → u11(ack_in(m, s(0)))
u11(ack_out(n)) → ack_out(n)
ack_in(s(m), s(n)) → u21(ack_in(s(m), n), m)
u21(ack_out(n), m) → u22(ack_in(m, n))
u22(ack_out(n)) → ack_out(n)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE