YES Termination w.r.t. Q proof of Rubio_04_elimdupl.ari

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

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

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 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:

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 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:

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 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:

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 transformation)

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 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:

  • 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 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.

(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 transformation)

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 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:

  • 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 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.

(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 transformation)

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 transformation)

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: 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

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

add_1=1
nil=1

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 transformation)

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

(29) YES