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

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)
+'(0(x), j(y)) -> +'(x, y)
+'(j(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(j(x), j(y)) -> +'(x, y)
+'(1(x), j(y)) -> 0'(+(x, y))
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), 1(y)) -> 0'(+(x, y))
+'(j(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
OPP(0(x)) -> 0'(opp(x))
OPP(0(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(j(x)) -> OPP(x)
-'(x, y) -> +'(x, opp(y))
-'(x, y) -> OPP(y)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(j(x), y) -> -'(0(*(x, y)), y)
*'(j(x), y) -> 0'(*(x, y))
*'(j(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

0(#) -> #
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), j(y)) -> 0(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(x, #) -> x
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(0(x), 0(y)) -> 0(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(#, 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(0(x₁)) = x₁

POL(1(x₁)) = 1 + x₁

POL(j(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, j, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳MRR

...

               →DP Problem 5

                 ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
0(#) -> #
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(0(x), 0(y)) -> 0(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(0(x), 1(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(#, x) -> x
+(1(x), 0(y)) -> 1(+(x, y))
+(+(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(0(x₁)) = 1 + x₁

POL(1(x₁)) = x₁

POL(j(x₁)) = x₁

POL(+(x₁, x₂)) = x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

We have the following set D of usable symbols: {#, 0, 1, j, +, +'}
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))
+(j(x), 0(y)) -> j(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(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 4

             ↳MRR

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳MRR

...

               →DP Problem 7

                 ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳MRR

...

               →DP Problem 8

                 ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

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:

+(x₁, x₂) -> +(x₁, x₂)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

OPP(j(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(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))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Strategy:

innermost

As we are in the innermost case, we can delete all 23 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 9

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

OPP(j(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(x)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

OPP(j(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(0(x)) -> OPP(x)

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

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

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

{1, 2, 3}	,	{1, 2, 3}
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)
j(x₁) -> j(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pairs:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 23 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 10

             ↳Size-Change Principle

Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

*'(*(x, y), z) -> *'(y, z)
*'(j(x), y) -> *'(x, y)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

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

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

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)
*(x₁, x₂) -> *(x₁, x₂)
j(x₁) -> j(x₁)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:03 minutes

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