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 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 UsableRulesProof (⇔)
    ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 UsableRulesProof (⇔)
    ↳22 QDP
    ↳23 QDPSizeChangeProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 PisEmptyProof (⇔)
    ↳29 TRUE

(0) Obligation:

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(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:

MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
MINUS(s(x), s(y)) → MINUS(x, any(y))
MINUS(s(x), s(y)) → ANY(y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → MINUS(max(x, y), min(x, y))
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
ANY(s(x)) → ANY(x)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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 5 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ANY(s(x)) → ANY(x)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

ANY(s(x)) → ANY(x)

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

(8) 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:

  • ANY(s(x)) → ANY(x)
    The graph contains the following edges 1 > 1

(9) TRUE

(10) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, any(y))

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, any(y))

The TRS R consists of the following rules:

any(s(x)) → s(s(any(x)))
any(x) → x

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

(13) 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:

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

(14) TRUE

(15) 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(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) 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.

(18) 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

(19) TRUE

(20) 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(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) 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.

(23) 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

(24) TRUE

(25) Obligation:

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

GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(GCD(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/01\
\00/
·x2

POL(s(x1)) =
/1\
\0/
 +
/11\
\11/
·x1

POL(minus(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2

POL(max(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/10\
\01/
·x2

POL(min(x1, x2)) =
/0\
\0/
 +
/10\
\01/
·x1 +
/00\
\00/
·x2

POL(0) =
/0\
\0/

POL(any(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

The following usable rules [FROCOS05] were oriented:

min(x, 0) → 0
min(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x

(27) Obligation:

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE