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

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

(0) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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:

-1(s(x), s(y)) → -1(x, y)
<=1(s(x), s(y)) → <=1(x, y)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
F(s(x), 0, z, u) → -1(z, s(x))
F(s(x), s(y), z, u) → IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) → <=1(x, y)
F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) → -1(y, x)
F(s(x), s(y), z, u) → F(x, u, z, u)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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 3 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

<=1(s(x), s(y)) → <=1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


<=1(s(x), s(y)) → <=1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
<=1(x1, x2)  =  <=1(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
<=^11: [1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Q is empty.
We have to consider all minimal (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.


-1(s(x), s(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  -1(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
-^11: [1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  x1
s(x1)  =  s(x1)
-(x1, x2)  =  x2
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
0 > s1

Status:
s1: [1]
0: []


The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  F(x1, x2, x3, x4)
s(x1)  =  s(x1)
-(x1, x2)  =  x1
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
F4 > s1
0 > s1

Status:
F4: [1,4,3,2]
s1: [1]
0: []


The following usable rules [FROCOS05] were oriented:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)

(19) Obligation:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE