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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE

(0) Obligation:

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

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:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(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:

MIN(s(X), s(Y)) → MIN(X, Y)
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) → MIN(X, Y)
LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) → QUOT(X, s(s(0)))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(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 3 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) 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) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.


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)  =  MIN(x2)
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
quot(x1, x2)  =  x1
log(x1)  =  log(x1)

Recursive Path Order [RPO].
Precedence:
log1 > [MIN1, s1] > 0


The following usable rules [FROCOS05] were oriented:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

(9) Obligation:

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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:

QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.


QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
quot(x1, x2)  =  x1
log(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
[QUOT2, s1]


The following usable rules [FROCOS05] were oriented:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

(14) Obligation:

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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:

LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))

The TRS R consists of the following rules:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.


LOG(s(s(X))) → LOG(s(quot(X, s(s(0)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LOG(x1)  =  LOG(x1)
s(x1)  =  s(x1)
quot(x1, x2)  =  x1
0  =  0
min(x1, x2)  =  x1
log(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
[LOG1, s1] > 0


The following usable rules [FROCOS05] were oriented:

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

(19) Obligation:

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

min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
log(s(0)) → 0
log(s(s(X))) → s(log(s(quot(X, s(s(0))))))

The set Q consists of the following terms:

min(x0, 0)
min(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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