Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
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(#)), #)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> 0'(+(+(x, y), 1(#)))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
-'(0(x), 0(y)) -> 0'(-(x, y))
-'(0(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(1(x), 1(y)) -> 0'(-(x, y))
-'(1(x), 1(y)) -> -'(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(x) -> -'(log'(x), 1(#))
LOG(x) -> LOG'(x)
LOG'(1(x)) -> +'(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)) -> +'(log'(x), 1(#))
LOG'(0(x)) -> LOG'(x)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)

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(#)), #)

Strategy:

innermost

As we are in the innermost case, we can delete all 21 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

Rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(#) = 0

POL(0(x₁)) = x₁

POL(1(x₁)) = 1 + x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 7

                 ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(0(x), 0(y)) -> +'(x, y)

Rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
0(#) -> #
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(#) = 0

POL(0(x₁)) = 1 + x₁

POL(1(x₁)) = x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = x₁ + x₂

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(0(x), 0(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

0(#) -> #
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 8

                 ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))

Rules:

+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(#) = 0

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

We have the following set D of usable symbols: {+, +'}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

+(x, #) -> x
+(#, x) -> x

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 9

                 ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))

Rule:

+(+(x, y), z) -> +(x, +(y, z))

Strategy:

innermost

We have the following set of usable rules:

+(+(x, y), z) -> +(x, +(y, z))

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(+(x₁, x₂)) = 1 + x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

We have the following set D of usable symbols: {+, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(+(x, y), z) -> +'(y, z)

No Rules can be deleted.

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

+'(+(x, y), z) -> +'(x, +(y, z))

Rule:

+(+(x, y), z) -> +(x, +(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 0(y)) -> -'(x, y)

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(#)), #)

Strategy:

innermost

As we are in the innermost case, we can delete all 22 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 11

             ↳Modular Removal of Rules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 0(y)) -> -'(x, y)

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

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

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(#) = 0

POL(-'(x₁, x₂)) = 1 + x₁ + x₂

POL(0(x₁)) = 1 + x₁

POL(1(x₁)) = 1 + x₁

POL(-(x₁, x₂)) = x₁ + x₂

We have the following set D of usable symbols: {#, -', 0, 1, -}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 0(y)) -> -'(x, y)

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

GE(#, 0(x)) -> GE(#, x)

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(#)), #)

Strategy:

innermost

As we are in the innermost case, we can delete all 29 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 12

             ↳Size-Change Principle

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

GE(#, 0(x)) -> GE(#, x)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

GE(#, 0(x)) -> GE(#, x)

and get the following Size-Change Graph(s):

{1}	,	{1}
1	=	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	=	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

0(x₁) -> 0(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳Usable Rules (Innermost)

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

GE(1(x), 1(y)) -> GE(x, y)
GE(1(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> GE(y, x)
GE(0(x), 0(y)) -> GE(x, y)

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(#)), #)

Strategy:

innermost

As we are in the innermost case, we can delete all 29 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 13

             ↳Size-Change Principle

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

GE(1(x), 1(y)) -> GE(x, y)
GE(1(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> GE(y, x)
GE(0(x), 0(y)) -> GE(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

GE(1(x), 1(y)) -> GE(x, y)
GE(1(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> GE(y, x)
GE(0(x), 0(y)) -> GE(x, y)

and get the following Size-Change Graph(s):

{1, 2, 3, 4}	,	{1, 2, 3, 4}
1	>	1
2	>	2

{1, 2, 3, 4}	,	{1, 2, 3, 4}
1	>	2
2	>	1

which lead(s) to this/these maximal multigraph(s):

{1, 2, 3, 4}	,	{1, 2, 3, 4}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳Usable Rules (Innermost)

Dependency Pairs:

LOG'(0(x)) -> LOG'(x)
LOG'(1(x)) -> LOG'(x)

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(#)), #)

Strategy:

innermost

As we are in the innermost case, we can delete all 29 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 14

             ↳Size-Change Principle

Dependency Pairs:

LOG'(0(x)) -> LOG'(x)
LOG'(1(x)) -> LOG'(x)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

LOG'(0(x)) -> LOG'(x)
LOG'(1(x)) -> LOG'(x)

and get the following Size-Change Graph(s):

{1, 2}	,	{1, 2}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1, 2}	,	{1, 2}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

0(x₁) -> 0(x₁)
1(x₁) -> 1(x₁)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes

POL(#)	= 0
POL(0(x₁))	= x₁
POL(1(x₁))	= 1 + x₁
POL(+(x₁, x₂))	= x₁ + x₂
POL(+'(x₁, x₂))	= 1 + x₁ + x₂

POL(#)	= 0
POL(-'(x₁, x₂))	= 1 + x₁ + x₂
POL(0(x₁))	= 1 + x₁
POL(1(x₁))	= 1 + x₁
POL(-(x₁, x₂))	= x₁ + x₂