Termination Proof using AProVE

Term Rewriting System R:
[x, y, n, m]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
APP(add(n, x), y) -> APP(x, y)
LOW(n, add(m, x)) -> IF_LOW(le(m, n), n, add(m, x))
LOW(n, add(m, x)) -> LE(m, n)
IF_LOW(true, n, add(m, x)) -> LOW(n, x)
IF_LOW(false, n, add(m, x)) -> LOW(n, x)
HIGH(n, add(m, x)) -> IF_HIGH(le(m, n), n, add(m, x))
HIGH(n, add(m, x)) -> LE(m, n)
IF_HIGH(true, n, add(m, x)) -> HIGH(n, x)
IF_HIGH(false, n, add(m, x)) -> HIGH(n, x)
QUICKSORT(add(n, x)) -> APP(quicksort(low(n, x)), add(n, quicksort(high(n, x))))
QUICKSORT(add(n, x)) -> QUICKSORT(low(n, x))
QUICKSORT(add(n, x)) -> LOW(n, x)
QUICKSORT(add(n, x)) -> QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) -> HIGH(n, x)

Furthermore, R contains seven SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 19 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 8

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 19 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 9

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳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

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

APP(add(n, x), y) -> APP(x, y)

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 19 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 10

             ↳Size-Change Principle

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

APP(add(n, x), y) -> APP(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

APP(add(n, x), y) -> APP(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:

add(x₁, x₂) -> add(x₁, x₂)

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

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 17 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 11

             ↳Negative Polynomial Order

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x₁, x₂) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( minus(x₁, x₂) ) = x₁

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 11

             ↳Neg POLO

...

               →DP Problem 12

                 ↳Dependency Graph

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pair:

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x

Strategy:

innermost

Using the Dependency Graph resulted in 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)

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pairs:

IF_LOW(false, n, add(m, x)) -> LOW(n, x)
IF_LOW(true, n, add(m, x)) -> LOW(n, x)
LOW(n, add(m, x)) -> IF_LOW(le(m, n), n, add(m, x))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 16 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 13

             ↳Size-Change Principle

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

Dependency Pairs:

IF_LOW(false, n, add(m, x)) -> LOW(n, x)
IF_LOW(true, n, add(m, x)) -> LOW(n, x)
LOW(n, add(m, x)) -> IF_LOW(le(m, n), n, add(m, x))

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_LOW(false, n, add(m, x)) -> LOW(n, x)
IF_LOW(true, n, add(m, x)) -> LOW(n, x)
LOW(n, add(m, x)) -> IF_LOW(le(m, n), n, add(m, x))

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

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

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

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

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

{3}	,	{1, 2}
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:

add(x₁, x₂) -> add(x₁, x₂)

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

         ↳UsableRules

       →DP Problem 6

         ↳Usable Rules (Innermost)

       →DP Problem 7

         ↳UsableRules

Dependency Pairs:

IF_HIGH(false, n, add(m, x)) -> HIGH(n, x)
IF_HIGH(true, n, add(m, x)) -> HIGH(n, x)
HIGH(n, add(m, x)) -> IF_HIGH(le(m, n), n, add(m, x))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 16 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 6

         ↳UsableRules

           →DP Problem 14

             ↳Size-Change Principle

       →DP Problem 7

         ↳UsableRules

Dependency Pairs:

IF_HIGH(false, n, add(m, x)) -> HIGH(n, x)
IF_HIGH(true, n, add(m, x)) -> HIGH(n, x)
HIGH(n, add(m, x)) -> IF_HIGH(le(m, n), n, add(m, x))

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_HIGH(false, n, add(m, x)) -> HIGH(n, x)
IF_HIGH(true, n, add(m, x)) -> HIGH(n, x)
HIGH(n, add(m, x)) -> IF_HIGH(le(m, n), n, add(m, x))

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

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

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

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

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

{1, 2}	,	{3}
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:

add(x₁, x₂) -> add(x₁, x₂)

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

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳Usable Rules (Innermost)

Dependency Pairs:

QUICKSORT(add(n, x)) -> QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) -> QUICKSORT(low(n, x))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
app(nil, y) -> y
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) -> add(m, low(n, x))
if_low(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
quicksort(nil) -> nil
quicksort(add(n, x)) -> app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Strategy:

innermost

As we are in the innermost case, we can delete all 8 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 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

           →DP Problem 15

             ↳Negative Polynomial Order

Dependency Pairs:

QUICKSORT(add(n, x)) -> QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) -> QUICKSORT(low(n, x))

Rules:

le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(false, n, add(m, x)) -> low(n, x)
if_low(true, n, add(m, x)) -> add(m, low(n, x))

Strategy:

innermost

The following Dependency Pairs can be strictly oriented using the given order.

QUICKSORT(add(n, x)) -> QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) -> QUICKSORT(low(n, x))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(false, n, add(m, x)) -> low(n, x)
if_low(true, n, add(m, x)) -> add(m, low(n, x))

Used ordering:
Polynomial Order with Interpretation:

POL( QUICKSORT(x₁) ) = x₁

POL( add(x₁, x₂) ) = x₂ + 1

POL( high(x₁, x₂) ) = x₂

POL( low(x₁, x₂) ) = x₂

POL( le(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

POL( if_high(x₁, ..., x₃) ) = x₃

POL( nil ) = 0

POL( if_low(x₁, ..., x₃) ) = x₃

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

       →DP Problem 6

         ↳UsableRules

       →DP Problem 7

         ↳UsableRules

           →DP Problem 15

             ↳Neg POLO

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pair:

Rules:

le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false
if_high(true, n, add(m, x)) -> high(n, x)
if_high(false, n, add(m, x)) -> add(m, high(n, x))
high(n, nil) -> nil
high(n, add(m, x)) -> if_high(le(m, n), n, add(m, x))
low(n, nil) -> nil
low(n, add(m, x)) -> if_low(le(m, n), n, add(m, x))
if_low(false, n, add(m, x)) -> low(n, x)
if_low(true, n, add(m, x)) -> add(m, low(n, x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:04 minutes