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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 QDP
    ↳19 UsableRulesProof (⇔)
    ↳20 QDP
    ↳21 QReductionProof (⇔)
    ↳22 QDP
    ↳23 QDPSizeChangeProof (⇔)
    ↳24 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

(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:

MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x)) → MINUS(s(x), g(f(x)))
F(s(x)) → G(f(x))
F(s(x)) → F(x)
G(s(x)) → MINUS(s(x), f(g(x)))
G(s(x)) → F(g(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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:

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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].

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

(11) Obligation:

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

MINUS(s(x), s(y)) → MINUS(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:

  • MINUS(s(x), s(y)) → MINUS(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:

F(s(x)) → G(f(x))
G(s(x)) → F(g(x))
F(s(x)) → F(x)
G(s(x)) → G(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


G(s(x)) → F(g(x))
F(s(x)) → F(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(F(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(s(x1)) =
/0\
\1/
 +
/11\
\11/
·x1

POL(G(x1)) =
/1\
\0/
 +
/10\
\00/
·x1

POL(f(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

POL(g(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

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

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

g(s(x)) → minus(s(x), f(g(x)))
g(0) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))

(16) Obligation:

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

F(s(x)) → G(f(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(19) 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.

(20) Obligation:

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

G(s(x)) → G(x)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

(21) 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].

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

(22) Obligation:

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

G(s(x)) → G(x)

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:

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

(24) TRUE