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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 AND
    ↳11 QDP
        ↳12 QDPOrderProof (⇔)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔)
        ↳15 TRUE
    ↳16 QDP
        ↳17 QDPOrderProof (⇔)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔)
        ↳20 TRUE

(0) Obligation:

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

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, 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:

ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
F(x, y) → ACK(x, y)
ACK(s(x), y) → F(x, x)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACK(x0, x1, x2)  =  ACK(x0)
F(x0, x1, x2)  =  F(x0, x1, x2)

Tags:
ACK has argument tags [2,5,5] and root tag 0
F has argument tags [2,2,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(ACK(x1, x2)) = x1   
POL(F(x1, x2)) = 1   
POL(ack(x1, x2)) = 1 + x1   
POL(f(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

ACK(s(x), s(y)) → ACK(s(x), y)
F(s(x), y) → F(x, s(x))
F(x, s(y)) → F(y, x)
F(x, y) → ACK(x, y)
ACK(s(x), y) → F(x, x)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, 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].


The following pairs can be oriented strictly and are deleted.


F(s(x), y) → F(x, s(x))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACK(x0, x1, x2)  =  ACK(x1, x2)
F(x0, x1, x2)  =  F(x0, x1, x2)

Tags:
ACK has argument tags [4,1,0] and root tag 0
F has argument tags [1,4,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACK(x1, x2)) = 1   
POL(F(x1, x2)) = 1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

ACK(s(x), s(y)) → ACK(s(x), y)
F(x, s(y)) → F(y, x)
F(x, y) → ACK(x, y)
ACK(s(x), y) → F(x, x)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACK(s(x), y) → F(x, x)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACK(x0, x1, x2)  =  ACK(x0, x1)
F(x0, x1, x2)  =  F(x1, x2)

Tags:
ACK has argument tags [0,0,4] and root tag 0
F has argument tags [0,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(ACK(x1, x2)) = 0   
POL(F(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

ACK(s(x), s(y)) → ACK(s(x), y)
F(x, s(y)) → F(y, x)
F(x, y) → ACK(x, y)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Complex Obligation (AND)

(11) Obligation:

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

ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACK(s(x), s(y)) → ACK(s(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACK(x0, x1, x2)  =  ACK(x2)

Tags:
ACK has argument tags [0,3,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) Obligation:

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

F(x, s(y)) → F(y, x)

The TRS R consists of the following rules:

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(x, s(y)) → F(y, x)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2)  =  F(x1, x2)

Tags:
F has argument tags [1,3,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
f(s(x), y) → f(x, s(x))
f(x, s(y)) → f(y, x)
f(x, y) → ack(x, y)
ack(s(x), y) → f(x, x)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE