YES Termination w.r.t. Q proof of CiME_04_log2.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 71 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 19 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 MRRProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 PisEmptyProof (⇔, 0 ms)
        ↳25 YES
    ↳26 QDP
        ↳27 UsableRulesProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 MRRProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 PisEmptyProof (⇔, 0 ms)
        ↳32 YES
    ↳33 QDP
        ↳34 UsableRulesProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 QDPSizeChangeProof (⇔, 0 ms)
        ↳37 YES

(0) Obligation:

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(#) → #
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(#) = 0   
POL(+(x1, x2)) = x1 + x2   
POL(-(x1, x2)) = x1 + x2   
POL(0(x1)) = 2·x1   
POL(1(x1)) = 2·x1   
POL(false) = 0   
POL(ge(x1, x2)) = 2·x1 + x2   
POL(if(x1, x2, x3)) = x1 + x2 + x3   
POL(log(x1)) = 1 + 2·x1   
POL(log'(x1)) = 1 + 2·x1   
POL(not(x1)) = x1   
POL(true) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

log'(#) → #


(2) Obligation:

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log(x) → -(log'(x), 1(#))
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(#) = 0   
POL(+(x1, x2)) = x1 + x2   
POL(-(x1, x2)) = x1 + x2   
POL(0(x1)) = 2·x1   
POL(1(x1)) = 2·x1   
POL(false) = 0   
POL(ge(x1, x2)) = x1 + x2   
POL(if(x1, x2, x3)) = 2·x1 + x2 + x3   
POL(log(x1)) = 2 + 2·x1   
POL(log'(x1)) = 1 + 2·x1   
POL(not(x1)) = 2·x1   
POL(true) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

log(x) → -(log'(x), 1(#))


(4) Obligation:

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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(x), 0(y)) → 01(+(x, y))
+1(0(x), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → 01(+(+(x, y), 1(#)))
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
+1(1(x), 1(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)
-1(0(x), 0(y)) → 01(-(x, y))
-1(0(x), 0(y)) → -1(x, y)
-1(0(x), 1(y)) → -1(-(x, y), 1(#))
-1(0(x), 1(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)
-1(1(x), 1(y)) → 01(-(x, y))
-1(1(x), 1(y)) → -1(x, y)
GE(0(x), 0(y)) → GE(x, y)
GE(0(x), 1(y)) → NOT(ge(y, x))
GE(0(x), 1(y)) → GE(y, x)
GE(1(x), 0(y)) → GE(x, y)
GE(1(x), 1(y)) → GE(x, y)
GE(#, 0(x)) → GE(#, x)
LOG'(1(x)) → +1(log'(x), 1(#))
LOG'(1(x)) → LOG'(x)
LOG'(0(x)) → IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
LOG'(0(x)) → GE(x, 1(#))
LOG'(0(x)) → +1(log'(x), 1(#))
LOG'(0(x)) → LOG'(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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 5 SCCs with 9 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

GE(#, 0(x)) → GE(#, x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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:

GE(#, 0(x)) → GE(#, x)

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:

  • GE(#, 0(x)) → GE(#, x)
    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:

GE(0(x), 1(y)) → GE(y, x)
GE(0(x), 0(y)) → GE(x, y)
GE(1(x), 0(y)) → GE(x, y)
GE(1(x), 1(y)) → GE(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(15) 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.

(16) Obligation:

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

GE(0(x), 1(y)) → GE(y, x)
GE(0(x), 0(y)) → GE(x, y)
GE(1(x), 0(y)) → GE(x, y)
GE(1(x), 1(y)) → GE(x, y)

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

(17) 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:

  • GE(0(x), 1(y)) → GE(y, x)
    The graph contains the following edges 2 > 1, 1 > 2

  • GE(0(x), 0(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • GE(1(x), 0(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • GE(1(x), 1(y)) → GE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(18) YES

(19) Obligation:

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

-1(0(x), 1(y)) → -1(-(x, y), 1(#))
-1(0(x), 1(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)
-1(1(x), 1(y)) → -1(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(20) 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.

(21) Obligation:

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

-1(0(x), 1(y)) → -1(-(x, y), 1(#))
-1(0(x), 1(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)
-1(1(x), 1(y)) → -1(x, y)

The TRS R consists of the following rules:

-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
0(#) → #

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

(22) 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(0(x), 1(y)) → -1(-(x, y), 1(#))
-1(0(x), 1(y)) → -1(x, y)
-1(0(x), 0(y)) → -1(x, y)
-1(1(x), 0(y)) → -1(x, y)
-1(1(x), 1(y)) → -1(x, y)

Strictly oriented rules of the TRS R:

-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
0(#) → #

Used ordering: Knuth-Bendix order [KBO] with precedence:
-2 > -^12 > 01 > 11 > #

and weight map:

#=2
0_1=5
1_1=3
-_2=0
-^1_2=0

The variable weight is 1

(23) Obligation:

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) YES

(26) Obligation:

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

+1(0(x), 1(y)) → +1(x, y)
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
+1(1(x), 1(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(27) 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.

(28) Obligation:

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

+1(0(x), 1(y)) → +1(x, y)
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
+1(1(x), 1(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)

The TRS R consists of the following rules:

+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
0(#) → #

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

(29) 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(0(x), 1(y)) → +1(x, y)
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
+1(1(x), 1(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)

Strictly oriented rules of the TRS R:

+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
0(#) → #

Used ordering: Knuth-Bendix order [KBO] with precedence:
# > +2 > 11 > 01 > +^12

and weight map:

#=2
0_1=1
1_1=3
+_2=0
+^1_2=0

The variable weight is 1

(30) Obligation:

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) YES

(33) Obligation:

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

LOG'(0(x)) → LOG'(x)
LOG'(1(x)) → LOG'(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
-(#, x) → #
-(x, #) → x
-(0(x), 0(y)) → 0(-(x, y))
-(0(x), 1(y)) → 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) → 1(-(x, y))
-(1(x), 1(y)) → 0(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(0(x), 0(y)) → ge(x, y)
ge(0(x), 1(y)) → not(ge(y, x))
ge(1(x), 0(y)) → ge(x, y)
ge(1(x), 1(y)) → ge(x, y)
ge(x, #) → true
ge(#, 0(x)) → ge(#, x)
ge(#, 1(x)) → false
log'(1(x)) → +(log'(x), 1(#))
log'(0(x)) → if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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

(34) 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.

(35) Obligation:

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

LOG'(0(x)) → LOG'(x)
LOG'(1(x)) → LOG'(x)

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

(36) 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:

  • LOG'(0(x)) → LOG'(x)
    The graph contains the following edges 1 > 1

  • LOG'(1(x)) → LOG'(x)
    The graph contains the following edges 1 > 1

(37) YES