YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Trafo_06/dup09.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 39 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 MNOCProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 MRRProof (⇔, 11 ms)
        ↳15 QDP
        ↳16 QDPOrderProof (⇔, 10 ms)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 MNOCProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 MRRProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 MRRProof (⇔, 3 ms)
        ↳28 QDP
        ↳29 PisEmptyProof (⇔, 0 ms)
        ↳30 YES
    ↳31 QDP
        ↳32 UsableRulesProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 QDPOrderProof (⇔, 26 ms)
        ↳35 QDP
        ↳36 PisEmptyProof (⇔, 0 ms)
        ↳37 YES
    ↳38 QDP
        ↳39 MRRProof (⇔, 58 ms)
        ↳40 QDP
        ↳41 MRRProof (⇔, 23 ms)
        ↳42 QDP
        ↳43 PisEmptyProof (⇔, 0 ms)
        ↳44 YES

(0) Obligation:

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

0(0(*(*(x)))) → *(*(1(1(x))))
1(1(*(*(x)))) → 0(0(#(#(x))))
#(#(0(0(x)))) → 0(0(#(#(x))))
#(#(1(1(x)))) → 1(1(#(#(x))))
#(#($($(x)))) → *(*($($(x))))
#(#(#(#(x)))) → #(#(x))
#(#(*(*(x)))) → *(*(x))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))
#(#(#(#(x)))) → #(#(x))
*(*(#(#(x)))) → *(*(x))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(#(x1)) = 1 + x1   
POL($(x1)) = x1   
POL(*(x1)) = 1 + x1   
POL(0(x1)) = x1   
POL(1(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

#(#(#(#(x)))) → #(#(x))
*(*(#(#(x)))) → *(*(x))


(4) Obligation:

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

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

*1(*(0(0(x)))) → 11(1(*(*(x))))
*1(*(0(0(x)))) → 11(*(*(x)))
*1(*(0(0(x)))) → *1(*(x))
*1(*(0(0(x)))) → *1(x)
*1(*(1(1(x)))) → 01(0(x))
*1(*(1(1(x)))) → 01(x)
01(0(#(#(x)))) → 01(0(x))
01(0(#(#(x)))) → 01(x)
11(1(#(#(x)))) → 11(1(x))
11(1(#(#(x)))) → 11(x)
$1($(#(#(x)))) → $1($(*(*(x))))
$1($(#(#(x)))) → $1(*(*(x)))
$1($(#(#(x)))) → *1(*(x))
$1($(#(#(x)))) → *1(x)

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

11(1(#(#(x)))) → 11(x)
11(1(#(#(x)))) → 11(1(x))

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

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

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

11(1(#(#(x)))) → 11(x)
11(1(#(#(x)))) → 11(1(x))

The TRS R consists of the following rules:

1(1(#(#(x)))) → #(#(1(1(x))))

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

(12) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(13) Obligation:

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

11(1(#(#(x)))) → 11(x)
11(1(#(#(x)))) → 11(1(x))

The TRS R consists of the following rules:

1(1(#(#(x)))) → #(#(1(1(x))))

The set Q consists of the following terms:

1(1(#(#(x0))))

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

(14) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

11(1(#(#(x)))) → 11(x)


Used ordering: Polynomial interpretation [POLO]:

POL(#(x1)) = x1   
POL(1(x1)) = 1 + 2·x1   
POL(11(x1)) = 2·x1   

(15) Obligation:

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

11(1(#(#(x)))) → 11(1(x))

The TRS R consists of the following rules:

1(1(#(#(x)))) → #(#(1(1(x))))

The set Q consists of the following terms:

1(1(#(#(x0))))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


11(1(#(#(x)))) → 11(1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( 11(x1) ) = max{0, 2x1 - 2}

POL( 1(x1) ) = x1

POL( #(x1) ) = 2x1 + 2


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

1(1(#(#(x)))) → #(#(1(1(x))))

(17) Obligation:

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

1(1(#(#(x)))) → #(#(1(1(x))))

The set Q consists of the following terms:

1(1(#(#(x0))))

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) YES

(20) Obligation:

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

01(0(#(#(x)))) → 01(x)
01(0(#(#(x)))) → 01(0(x))

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

01(0(#(#(x)))) → 01(x)
01(0(#(#(x)))) → 01(0(x))

The TRS R consists of the following rules:

0(0(#(#(x)))) → #(#(0(0(x))))

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

(23) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(24) Obligation:

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

01(0(#(#(x)))) → 01(x)
01(0(#(#(x)))) → 01(0(x))

The TRS R consists of the following rules:

0(0(#(#(x)))) → #(#(0(0(x))))

The set Q consists of the following terms:

0(0(#(#(x0))))

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

(25) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

01(0(#(#(x)))) → 01(x)


Used ordering: Polynomial interpretation [POLO]:

POL(#(x1)) = x1   
POL(0(x1)) = 1 + 2·x1   
POL(01(x1)) = 2·x1   

(26) Obligation:

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

01(0(#(#(x)))) → 01(0(x))

The TRS R consists of the following rules:

0(0(#(#(x)))) → #(#(0(0(x))))

The set Q consists of the following terms:

0(0(#(#(x0))))

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

(27) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

01(0(#(#(x)))) → 01(0(x))

Strictly oriented rules of the TRS R:

0(0(#(#(x)))) → #(#(0(0(x))))

Used ordering: Polynomial interpretation [POLO]:

POL(#(x1)) = 3 + 3·x1   
POL(0(x1)) = 3·x1   
POL(01(x1)) = 2·x1   

(28) Obligation:

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

0(0(#(#(x0))))

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) YES

(31) Obligation:

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

*1(*(0(0(x)))) → *1(x)
*1(*(0(0(x)))) → *1(*(x))

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

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

(32) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(33) Obligation:

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

*1(*(0(0(x)))) → *1(x)
*1(*(0(0(x)))) → *1(*(x))

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


*1(*(0(0(x)))) → *1(x)
*1(*(0(0(x)))) → *1(*(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( *1(x1) ) = max{0, 2x1 - 2}

POL( *(x1) ) = x1 + 2

POL( 0(x1) ) = 2x1 + 1

POL( 1(x1) ) = 1

POL( #(x1) ) = 1


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

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))

(35) Obligation:

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

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))

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

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) YES

(38) Obligation:

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

$1($(#(#(x)))) → $1(*(*(x)))
$1($(#(#(x)))) → $1($(*(*(x))))

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

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

(39) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

$1($(#(#(x)))) → $1(*(*(x)))


Used ordering: Polynomial interpretation [POLO]:

POL(#(x1)) = x1   
POL($(x1)) = 1 + 2·x1   
POL($1(x1)) = 2·x1   
POL(*(x1)) = x1   
POL(0(x1)) = 2·x1   
POL(1(x1)) = 2·x1   

(40) Obligation:

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

$1($(#(#(x)))) → $1($(*(*(x))))

The TRS R consists of the following rules:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

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

(41) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

$1($(#(#(x)))) → $1($(*(*(x))))

Strictly oriented rules of the TRS R:

*(*(0(0(x)))) → 1(1(*(*(x))))
*(*(1(1(x)))) → #(#(0(0(x))))
0(0(#(#(x)))) → #(#(0(0(x))))
1(1(#(#(x)))) → #(#(1(1(x))))
$($(#(#(x)))) → $($(*(*(x))))

Used ordering: Polynomial interpretation [POLO]:

POL(#(x1)) = 2 + 2·x1   
POL($(x1)) = 2·x1   
POL($1(x1)) = 2·x1   
POL(*(x1)) = 2·x1   
POL(0(x1)) = 2 + 3·x1   
POL(1(x1)) = 3 + 3·x1   

(42) Obligation:

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) YES