Termination Proof using AProVE

Term Rewriting System R:
[x, y]
app(p, 0) -> 0
app(p, app(s, x)) -> x
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
app(app(app(if, true), x), y) -> 0
app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, x)), y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) -> APP(le, x)
APP(app(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)
APP(app(minus, x), y) -> APP(app(if, app(app(le, x), y)), x)
APP(app(minus, x), y) -> APP(if, app(app(le, x), y))
APP(app(minus, x), y) -> APP(app(le, x), y)
APP(app(minus, x), y) -> APP(le, x)
APP(app(app(if, false), x), y) -> APP(s, app(app(minus, app(p, x)), y))
APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)
APP(app(app(if, false), x), y) -> APP(minus, app(p, x))
APP(app(app(if, false), x), y) -> APP(p, x)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

APP(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)

Rules:

app(p, 0) -> 0
app(p, app(s, x)) -> x
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
app(app(app(if, true), x), y) -> 0
app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, x)), y))

Strategy:

innermost

As we are in the innermost case, we can delete all 8 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳A-Transformation

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

APP(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)

Rule:

none

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳ATrans

...

               →DP Problem 4

                 ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

LE(s(x), s(y)) -> LE(x, y)

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:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

Dependency Pairs:

APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)
APP(app(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)

Rules:

app(p, 0) -> 0
app(p, app(s, x)) -> x
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(minus, x), y) -> app(app(app(if, app(app(le, x), y)), x), y)
app(app(app(if, true), x), y) -> 0
app(app(app(if, false), x), y) -> app(s, app(app(minus, app(p, x)), y))

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳A-Transformation

Dependency Pairs:

APP(app(app(if, false), x), y) -> APP(app(minus, app(p, x)), y)
APP(app(minus, x), y) -> APP(app(app(if, app(app(le, x), y)), x), y)

Rules:

app(p, 0) -> 0
app(p, app(s, x)) -> x
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

IF(false, x, y) -> MINUS(p(x), y)
MINUS(x, y) -> IF(le(x, y), x, y)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(x, y) -> IF(le(x, y), x, y)

three new Dependency Pairs are created:

MINUS(0, y'') -> IF(true, 0, y'')
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, x, y) -> MINUS(p(x), y)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF(false, x, y) -> MINUS(p(x), y)

two new Dependency Pairs are created:

IF(false, 0, y) -> MINUS(0, y)
IF(false, s(x''), y) -> MINUS(x'', y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 8

                 ↳Instantiation Transformation

Dependency Pairs:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, s(x''), y) -> MINUS(x'', y)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF(false, s(x''), y) -> MINUS(x'', y)

two new Dependency Pairs are created:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 9

                 ↳Usable Rules (Innermost)

Dependency Pairs:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 11

                 ↳Size-Change Principle

Dependency Pairs:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)

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

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

{2}	,	{2}
1	=	2
2	=	3

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

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

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

D_P: empty set
Oriented Rules: none

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

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 10

                 ↳Usable Rules (Innermost)

Dependency Pairs:

IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 12

                 ↳Size-Change Principle

Dependency Pairs:

IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

We number the DPs as follows:

IF(false, s(x''''), s(y'''')) -> MINUS(x'''', s(y''''))
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

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

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

{2}	,	{2}
1	=	2
2	=	3

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

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

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

D_P: empty set
Oriented Rules: none

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

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:06 minutes