Term Rewriting System R:
[x, y, u, z]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
<='(s(x), s(y)) -> <='(x, y)
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), 0, z, u) -> -'(z, s(x))
F(s(x), s(y), z, u) -> IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) -> <='(x, y)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) -> -'(y, x)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Usable Rules (Innermost)
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))


Strategy:

innermost




As we are in the innermost case, we can delete all 13 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 4
Size-Change Principle
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. -'(s(x), s(y)) -> -'(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

DP: 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.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)
       →DP Problem 3
UsableRules


Dependency Pair:

<='(s(x), s(y)) -> <='(x, y)


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))


Strategy:

innermost




As we are in the innermost case, we can delete all 13 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 5
Size-Change Principle
       →DP Problem 3
UsableRules


Dependency Pair:

<='(s(x), s(y)) -> <='(x, y)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. <='(s(x), s(y)) -> <='(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

DP: 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.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
Usable Rules (Innermost)


Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Negative Polynomial Order


Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)


Rules:


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


Strategy:

innermost




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

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)


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

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


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x4) ) = x1

POL( s(x1) ) = x1 + 1

POL( -(x1, x2) ) = x1


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Neg POLO
             ...
               →DP Problem 7
Negative Polynomial Order


Dependency Pair:

F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)


Rules:


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


Strategy:

innermost




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

F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)


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

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


Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x4) ) = x2

POL( s(x1) ) = x1 + 1

POL( -(x1, x2) ) = x1


This results in one new DP problem.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 6
Neg POLO
             ...
               →DP Problem 8
Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes