Term Rewriting System R:
[X, Y, X1, X2, X3, Z]
aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AAND(true, X) -> MARK(X)
AIF(true, X, Y) -> MARK(X)
AIF(false, X, Y) -> MARK(Y)
AADD(0, X) -> MARK(X)
MARK(and(X1, X2)) -> AAND(mark(X1), X2)
MARK(and(X1, X2)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)
MARK(add(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> AFIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) -> MARK(X1)
MARK(first(X1, X2)) -> MARK(X2)
MARK(from(X)) -> AFROM(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

AIF(false, X, Y) -> MARK(Y)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
AIF(true, X, Y) -> MARK(X)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)
MARK(and(X1, X2)) -> MARK(X1)
MARK(and(X1, X2)) -> AAND(mark(X1), X2)
AAND(true, X) -> MARK(X)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), X2, X3)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(and(x1, x2))=  x1 + x2  
  POL(MARK(x1))=  x1  
  POL(a__and(x1, x2))=  0  
  POL(false)=  0  
  POL(true)=  0  
  POL(mark(x1))=  0  
  POL(a__add(x1, x2))=  0  
  POL(a__from(x1))=  0  
  POL(a__first(x1, x2))=  0  
  POL(add(x1, x2))=  x1 + x2  
  POL(if(x1, x2, x3))=  1 + x1 + x2 + x3  
  POL(first(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(A__ADD(x1, x2))=  x2  
  POL(A__AND(x1, x2))=  x2  
  POL(cons(x1, x2))=  0  
  POL(a__if(x1, x2, x3))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  
  POL(A__IF(x1, x2, x3))=  x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

AIF(false, X, Y) -> MARK(Y)
MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)
AIF(true, X, Y) -> MARK(X)
MARK(and(X1, X2)) -> MARK(X1)
MARK(and(X1, X2)) -> AAND(mark(X1), X2)
AAND(true, X) -> MARK(X)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)
MARK(and(X1, X2)) -> MARK(X1)
AAND(true, X) -> MARK(X)
MARK(and(X1, X2)) -> AAND(mark(X1), X2)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(and(X1, X2)) -> MARK(X1)
MARK(and(X1, X2)) -> AAND(mark(X1), X2)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(and(x1, x2))=  1 + x1 + x2  
  POL(MARK(x1))=  x1  
  POL(a__and(x1, x2))=  0  
  POL(false)=  0  
  POL(true)=  0  
  POL(mark(x1))=  0  
  POL(a__add(x1, x2))=  0  
  POL(a__from(x1))=  0  
  POL(a__first(x1, x2))=  0  
  POL(add(x1, x2))=  x1 + x2  
  POL(if(x1, x2, x3))=  0  
  POL(first(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(A__ADD(x1, x2))=  x2  
  POL(A__AND(x1, x2))=  x2  
  POL(cons(x1, x2))=  0  
  POL(a__if(x1, x2, x3))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)
AAND(true, X) -> MARK(X)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)
MARK(first(X1, X2)) -> MARK(X2)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(add(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> AADD(mark(X1), X2)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(and(x1, x2))=  0  
  POL(MARK(x1))=  x1  
  POL(a__and(x1, x2))=  0  
  POL(false)=  0  
  POL(true)=  0  
  POL(mark(x1))=  0  
  POL(a__add(x1, x2))=  0  
  POL(a__from(x1))=  0  
  POL(a__first(x1, x2))=  0  
  POL(add(x1, x2))=  1 + x1 + x2  
  POL(if(x1, x2, x3))=  0  
  POL(first(x1, x2))=  x1 + x2  
  POL(0)=  0  
  POL(A__ADD(x1, x2))=  x2  
  POL(cons(x1, x2))=  0  
  POL(a__if(x1, x2, x3))=  0  
  POL(nil)=  0  
  POL(s(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X1)
AADD(0, X) -> MARK(X)
MARK(first(X1, X2)) -> MARK(X2)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pairs:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(first(X1, X2)) -> MARK(X2)
MARK(first(X1, X2)) -> MARK(X1)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MARK(x1))=  x1  
  POL(first(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


aand(true, X) -> mark(X)
aand(false, Y) -> false
aand(X1, X2) -> and(X1, X2)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
aadd(0, X) -> mark(X)
aadd(s(X), Y) -> s(add(X, Y))
aadd(X1, X2) -> add(X1, X2)
afirst(0, X) -> nil
afirst(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
afirst(X1, X2) -> first(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
mark(and(X1, X2)) -> aand(mark(X1), X2)
mark(if(X1, X2, X3)) -> aif(mark(X1), X2, X3)
mark(add(X1, X2)) -> aadd(mark(X1), X2)
mark(first(X1, X2)) -> afirst(mark(X1), mark(X2))
mark(from(X)) -> afrom(X)
mark(true) -> true
mark(false) -> false
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:05 minutes