YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 1 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 1 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPOrderProof [EQUIVALENT, 33 ms]
        (27) QDP
        (28) PisEmptyProof [EQUIVALENT, 0 ms]
        (29) YES


----------------------------------------

(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) Overlay + Local Confluence (EQUIVALENT)
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:

   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)))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
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:

   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)))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT)
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:

   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)))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT)
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:

   EQ(s(X), s(Y)) -> EQ(X, Y)

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

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))


----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(s(X), s(Y)) -> EQ(X, Y)

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

(12) QDPSizeChangeProof (EQUIVALENT)
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:
*EQ(s(X), s(Y)) -> EQ(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


----------------------------------------

(13)
YES

----------------------------------------

(14)
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)))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(16)
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)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))


----------------------------------------

(18)
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)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))

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

(19) QDPSizeChangeProof (EQUIVALENT)
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:
*RM(N, add(M, X)) -> IFRM(eq(N, M), N, add(M, X))
The graph contains the following edges 1 >= 2, 2 >= 3


*IFRM(true, N, add(M, X)) -> RM(N, X)
The graph contains the following edges 2 >= 1, 3 > 2


*IFRM(false, N, add(M, X)) -> RM(N, X)
The graph contains the following edges 2 >= 1, 3 > 2


----------------------------------------

(20)
YES

----------------------------------------

(21)
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)))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(22) UsableRulesProof (EQUIVALENT)
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.
----------------------------------------

(23)
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:

   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)
   eq(0, 0) -> true
   eq(0, s(X)) -> false
   eq(s(X), 0) -> false
   eq(s(X), s(Y)) -> eq(X, Y)
   ifrm(false, N, add(M, X)) -> add(M, rm(N, X))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))
   purge(nil)
   purge(add(x0, x1))

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

(24) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   purge(nil)
   purge(add(x0, x1))


----------------------------------------

(25)
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:

   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)
   eq(0, 0) -> true
   eq(0, s(X)) -> false
   eq(s(X), 0) -> false
   eq(s(X), s(Y)) -> eq(X, Y)
   ifrm(false, N, add(M, X)) -> add(M, rm(N, X))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))

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

(26) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


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:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( PURGE_1(x_1) ) = 2x_1 + 2
POL( rm_2(x_1, x_2) ) = x_2
POL( nil ) = 2
POL( add_2(x_1, x_2) ) = 2x_2 + 2
POL( ifrm_3(x_1, ..., x_3) ) = x_3
POL( eq_2(x_1, x_2) ) = 0
POL( true ) = 2
POL( 0 ) = 0
POL( s_1(x_1) ) = 2x_1
POL( false ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] 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))


----------------------------------------

(27)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   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)
   eq(0, 0) -> true
   eq(0, s(X)) -> false
   eq(s(X), 0) -> false
   eq(s(X), s(Y)) -> eq(X, Y)
   ifrm(false, N, add(M, X)) -> add(M, rm(N, X))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   rm(x0, nil)
   rm(x0, add(x1, x2))
   ifrm(true, x0, add(x1, x2))
   ifrm(false, x0, add(x1, x2))

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

(28) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(29)
YES