(0) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
min(0, y) → 0

The TRS R 2 is

f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The signature Sigma is {f}

(2) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
+1(s(x), y) → +1(x, y)
-1(s(x), s(y)) → -1(x, y)
*1(x, s(y)) → +1(x, *(x, y))
*1(x, s(y)) → *1(x, y)
F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))
F(s(x), s(y)) → -1(min(s(x), s(y)), max(s(x), s(y)))
F(s(x), s(y)) → MIN(s(x), s(y))
F(s(x), s(y)) → MAX(s(x), s(y))
F(s(x), s(y)) → *1(s(x), s(y))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) 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:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(8) 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)  =  x1
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
max(x1, x2)  =  max(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  x1
*(x1, x2)  =  *(x1, x2)
p(x1)  =  p(x1)
f(x1, x2)  =  f

Lexicographic Path Order [LPO].
Precedence:
max2 > s1 > [0, f]
*2 > +2 > s1 > [0, f]
p1 > [0, f]


The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

(9) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(s(x), 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)  =  x1
s(x1)  =  s(x1)
min(x1, x2)  =  min(x1, x2)
0  =  0
max(x1, x2)  =  max(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  x1
*(x1, x2)  =  *(x1, x2)
p(x1)  =  x1
f(x1, x2)  =  f

Lexicographic Path Order [LPO].
Precedence:
f > min2 > s1
f > min2 > 0
f > max2 > s1
f > *2 > 0
f > *2 > +2 > s1


The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

(14) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

*1(x, s(y)) → *1(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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.


*1(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)  =  x2
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
max(x1, x2)  =  max(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  x1
*(x1, x2)  =  *(x1, x2)
p(x1)  =  x1
f(x1, x2)  =  x1

Lexicographic Path Order [LPO].
Precedence:
max2 > s1 > 0
*2 > +2 > s1 > 0


The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

(19) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(22) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MAX(s(x), s(y)) → MAX(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MAX(x1, x2)  =  x1
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
max(x1, x2)  =  max(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  x1
*(x1, x2)  =  *(x1, x2)
p(x1)  =  x1
f(x1, x2)  =  f

Lexicographic Path Order [LPO].
Precedence:
max2 > [s1, f] > 0
*2 > +2 > [s1, f] > 0


The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

(24) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MIN(s(x), s(y)) → MIN(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2)  =  x1
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
max(x1, x2)  =  max(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  x1
*(x1, x2)  =  *(x1, x2)
p(x1)  =  p(x1)
f(x1, x2)  =  f

Lexicographic Path Order [LPO].
Precedence:
max2 > s1 > [0, f]
*2 > +2 > s1 > [0, f]
p1 > [0, f]


The following usable rules [FROCOS05] were oriented:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

(29) Obligation:

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

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The TRS R consists of the following rules:

min(0, y) → 0
min(x, 0) → 0
min(s(x), s(y)) → s(min(x, y))
max(0, y) → y
max(x, 0) → x
max(s(x), s(y)) → s(max(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(x, *(x, y))
p(s(x)) → x
f(s(x), s(y)) → f(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y)))

The set Q consists of the following terms:

min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
max(0, x0)
max(x0, 0)
max(s(x0), s(x1))
+(0, x0)
+(s(x0), x1)
-(x0, 0)
-(s(x0), s(x1))
*(x0, 0)
*(x0, s(x1))
p(s(x0))
f(s(x0), s(x1))

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