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

(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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

-¹(s(x), s(y)) → -¹(x, y)

R is empty.
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) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(11) Obligation:

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

-¹(s(x), s(y)) → -¹(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• -¹(s(x), s(y)) → -¹(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) 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.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

+¹(s(x), y) → +¹(x, y)

R is empty.
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.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(18) Obligation:

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

+¹(s(x), y) → +¹(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• +¹(s(x), y) → +¹(x, y)
  The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) 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.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

*¹(x, s(y)) → *¹(x, y)

R is empty.
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.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(25) Obligation:

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

*¹(x, s(y)) → *¹(x, y)

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• *¹(x, s(y)) → *¹(x, y)
  The graph contains the following edges 1 >= 1, 2 > 2

(27) TRUE

(28) 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.

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

R is empty.
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.

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(32) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

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

(33) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• MAX(s(x), s(y)) → MAX(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(34) TRUE

(35) 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.

(36) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(37) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

R is empty.
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.

(38) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(39) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

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

(40) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• MIN(s(x), s(y)) → MIN(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(41) TRUE

(42) 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.

(43) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

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

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.

(45) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

p(s(x0))
f(s(x0), s(x1))

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

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))

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

(47) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(min(s(x), s(y)), max(s(x), s(y))), *(s(x), s(y))) at position [0,0] we obtained the following new rules [LPAR04]:

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

(48) Obligation:

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

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

The TRS R consists of the following rules:

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

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))

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

(49) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(s(min(x, y)), max(s(x), s(y))), *(s(x), s(y))) at position [0,1] we obtained the following new rules [LPAR04]:

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

(50) Obligation:

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

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

The TRS R consists of the following rules:

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

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))

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

(51) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(s(min(x, y)), s(max(x, y))), *(s(x), s(y))) at position [0] we obtained the following new rules [LPAR04]:

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

(52) Obligation:

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

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

The TRS R consists of the following rules:

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

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))

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

(53) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(min(x, y), max(x, y)), *(s(x), s(y))) at position [1] we obtained the following new rules [LPAR04]:

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

(54) Obligation:

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

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

The TRS R consists of the following rules:

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

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))

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

(55) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(x), s(y)) → F(-(min(x, y), max(x, y)), +(s(x), *(s(x), y))) at position [1] we obtained the following new rules [LPAR04]:

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

(56) Obligation:

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

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

The TRS R consists of the following rules:

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

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))

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

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(*(x₁, x₂)) = x₂   
POL(+(x₁, x₂)) = 0   
POL(-(x₁, x₂)) = x₁   
POL(0) = 0   
POL(F(x₁, x₂)) = x₁   
POL(max(x₁, x₂)) = x₁ + x₂   
POL(min(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

-(s(x), s(y)) → -(x, y)
-(x, 0) → x
min(s(x), s(y)) → s(min(x, y))
min(x, 0) → 0
min(0, y) → 0

(58) Obligation:

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

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

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))

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

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) TRUE