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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳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

(0) Obligation:

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

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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:

A__AND(tt, X) → MARK(X)
A__PLUS(N, 0) → MARK(N)
A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → MARK(N)
A__PLUS(N, s(M)) → MARK(M)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__PLUS(N, 0) → MARK(N)
A__PLUS(N, s(M)) → MARK(N)
A__PLUS(N, s(M)) → MARK(M)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__AND(x1, x2)  =  x2
MARK(x1)  =  x1
A__PLUS(x1, x2)  =  A__PLUS(x1, x2)
0  =  0
s(x1)  =  x1
mark(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
plus(x1, x2)  =  plus(x1, x2)
tt  =  tt
a__plus(x1, x2)  =  a__plus(x1, x2)
a__and(x1, x2)  =  a__and(x1, x2)

Recursive Path Order [RPO].
Precedence:
[and2, aand2]
[plus2, aplus2] > APLUS2


The following usable rules [FROCOS05] were oriented:

mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__plus(N, 0) → mark(N)
a__and(tt, X) → mark(X)
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

(4) Obligation:

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

A__AND(tt, X) → MARK(X)
A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

Q is empty.
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 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

Q is empty.
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.


MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

Q is empty.
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:

A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

Q is empty.
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.


A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__PLUS(x1, x2)  =  x2
s(x1)  =  s(x1)
mark(x1)  =  x1
tt  =  tt
0  =  0
a__plus(x1, x2)  =  a__plus(x1, x2)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  x2
a__and(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
[aplus2, plus2] > s1


The following usable rules [FROCOS05] were oriented:

mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
a__plus(N, 0) → mark(N)
a__and(tt, X) → mark(X)
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

(14) Obligation:

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

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)

Q is empty.
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