Term Rewriting System R:
[X, Y, X1, X2, X3]
af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AF(X) -> AIF(mark(X), c, f(true))
AF(X) -> MARK(X)
AIF(true, X, Y) -> MARK(X)
AIF(false, X, Y) -> MARK(Y)
MARK(f(X)) -> AF(mark(X))
MARK(f(X)) -> MARK(X)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), mark(X2), X3)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> MARK(X2)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

MARK(if(X1, X2, X3)) -> MARK(X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
AIF(false, X, Y) -> MARK(Y)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), mark(X2), X3)
MARK(f(X)) -> MARK(X)
AF(X) -> MARK(X)
MARK(f(X)) -> AF(mark(X))
AIF(true, X, Y) -> MARK(X)
AF(X) -> AIF(mark(X), c, f(true))


Rules:


af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false


Strategy:

innermost




The following dependency pair can be strictly oriented:

AIF(false, X, Y) -> MARK(Y)


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false
af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(c)=  0  
  POL(MARK(x1))=  x1  
  POL(false)=  1  
  POL(a__if(x1, x2, x3))=  x1 + x2 + x3  
  POL(true)=  0  
  POL(A__F(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(f(x1))=  x1  
  POL(a__f(x1))=  x1  
  POL(A__IF(x1, x2, x3))=  x1 + x2 + x3  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polynomial Ordering


Dependency Pairs:

MARK(if(X1, X2, X3)) -> MARK(X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), mark(X2), X3)
MARK(f(X)) -> MARK(X)
AF(X) -> MARK(X)
MARK(f(X)) -> AF(mark(X))
AIF(true, X, Y) -> MARK(X)
AF(X) -> AIF(mark(X), c, f(true))


Rules:


af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false


Strategy:

innermost




The following dependency pairs can be strictly oriented:

MARK(f(X)) -> MARK(X)
MARK(f(X)) -> AF(mark(X))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false
af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(c)=  0  
  POL(MARK(x1))=  x1  
  POL(false)=  0  
  POL(a__if(x1, x2, x3))=  x1 + x2 + x3  
  POL(true)=  0  
  POL(A__F(x1))=  x1  
  POL(mark(x1))=  x1  
  POL(f(x1))=  1 + x1  
  POL(a__f(x1))=  1 + x1  
  POL(A__IF(x1, x2, x3))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pairs:

MARK(if(X1, X2, X3)) -> MARK(X2)
MARK(if(X1, X2, X3)) -> MARK(X1)
MARK(if(X1, X2, X3)) -> AIF(mark(X1), mark(X2), X3)
AF(X) -> MARK(X)
AIF(true, X, Y) -> MARK(X)
AF(X) -> AIF(mark(X), c, f(true))


Rules:


af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false


Strategy:

innermost




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


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Polo
             ...
               →DP Problem 4
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


af(X) -> aif(mark(X), c, f(true))
af(X) -> f(X)
aif(true, X, Y) -> mark(X)
aif(false, X, Y) -> mark(Y)
aif(X1, X2, X3) -> if(X1, X2, X3)
mark(f(X)) -> af(mark(X))
mark(if(X1, X2, X3)) -> aif(mark(X1), mark(X2), X3)
mark(c) -> c
mark(true) -> true
mark(false) -> false


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:01 minutes