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 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 DependencyGraphProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE

(0) Obligation:

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

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, 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:

EQ(s(X), s(Y)) → EQ(X, Y)
RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))
RM(N, add(M, X)) → EQ(N, M)
IFRM(true, N, add(M, X)) → RM(N, X)
IFRM(false, N, add(M, X)) → RM(N, X)
PURGE(add(N, X)) → PURGE(rm(N, X))
PURGE(add(N, X)) → RM(N, X)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, 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 3 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

EQ(s(X), s(Y)) → EQ(X, Y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))
IFRM(true, N, add(M, X)) → RM(N, X)
IFRM(false, N, add(M, X)) → RM(N, X)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


IFRM(true, N, add(M, X)) → RM(N, X)
IFRM(false, N, add(M, X)) → RM(N, X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
RM(x1, x2)  =  x2
add(x1, x2)  =  add(x1, x2)
IFRM(x1, x2, x3)  =  x3
eq(x1, x2)  =  x2
true  =  true
false  =  false
0  =  0
s(x1)  =  s

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(14) TRUE

(15) Obligation:

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

PURGE(add(N, X)) → PURGE(rm(N, X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PURGE(add(N, X)) → PURGE(rm(N, X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PURGE(x1)  =  x1
add(x1, x2)  =  add(x2)
rm(x1, x2)  =  x2
nil  =  nil
ifrm(x1, x2, x3)  =  x3
eq(x1, x2)  =  eq(x2)
true  =  true
false  =  false
0  =  0
s(x1)  =  s

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:

rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))

(17) Obligation:

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

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE