YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__f#(X) -> mark#(X)
p3: a__if#(true(),X,Y) -> mark#(X)
p4: a__if#(false(),X,Y) -> mark#(Y)
p5: mark#(f(X)) -> a__f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p8: mark#(if(X1,X2,X3)) -> mark#(X1)
p9: mark#(if(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(false(),X,Y) -> mark#(Y)
p3: mark#(if(X1,X2,X3)) -> mark#(X2)
p4: mark#(if(X1,X2,X3)) -> mark#(X1)
p5: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p6: a__if#(true(),X,Y) -> mark#(X)
p7: mark#(f(X)) -> mark#(X)
p8: mark#(f(X)) -> a__f#(mark(X))
p9: a__f#(X) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__f#_A(x1) = max{68, x1 + 26}
        a__if#_A(x1,x2,x3) = max{68, x2 + 31, x3 + 24}
        mark_A(x1) = x1 + 25
        c_A = 0
        f_A(x1) = x1 + 44
        true_A = 0
        false_A = 17
        mark#_A(x1) = max{68, x1 + 23}
        if_A(x1,x2,x3) = max{46, x1 + 18, x2 + 33, x3 + 25}
        a__f_A(x1) = max{69, x1 + 44}
        a__if_A(x1,x2,x3) = max{68, x1 + 18, x2 + 33, x3 + 25}
  
    precedence:
    
      a__f# = a__if# = mark = c = f = true = false = mark# = if = a__f = a__if
    
    partial status:
  
      pi(a__f#) = []
      pi(a__if#) = []
      pi(mark) = []
      pi(c) = []
      pi(f) = []
      pi(true) = []
      pi(false) = []
      pi(mark#) = []
      pi(if) = []
      pi(a__f) = []
      pi(a__if) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(false(),X,Y) -> mark#(Y)
p3: mark#(if(X1,X2,X3)) -> mark#(X1)
p4: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p5: a__if#(true(),X,Y) -> mark#(X)
p6: mark#(f(X)) -> mark#(X)
p7: mark#(f(X)) -> a__f#(mark(X))
p8: a__f#(X) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(true(),X,Y) -> mark#(X)
p3: mark#(f(X)) -> a__f#(mark(X))
p4: a__f#(X) -> mark#(X)
p5: mark#(f(X)) -> mark#(X)
p6: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p7: a__if#(false(),X,Y) -> mark#(Y)
p8: mark#(if(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__f#_A(x1) = max{11, x1 - 2}
        a__if#_A(x1,x2,x3) = max{11, x1 - 7, x2 - 2, x3 - 2}
        mark_A(x1) = x1 + 4
        c_A = 0
        f_A(x1) = x1 + 10
        true_A = 0
        mark#_A(x1) = max{11, x1 - 2}
        if_A(x1,x2,x3) = max{x1 + 6, x2 + 14, x3}
        false_A = 19
        a__f_A(x1) = max{14, x1 + 10}
        a__if_A(x1,x2,x3) = max{x1 + 6, x2 + 14, x3 + 4}
  
    precedence:
    
      a__f# = a__if# = c = mark# = false > mark = f = true = a__f = a__if > if
    
    partial status:
  
      pi(a__f#) = []
      pi(a__if#) = []
      pi(mark) = []
      pi(c) = []
      pi(f) = []
      pi(true) = []
      pi(mark#) = []
      pi(if) = [1]
      pi(false) = []
      pi(a__f) = []
      pi(a__if) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(true(),X,Y) -> mark#(X)
p3: mark#(f(X)) -> a__f#(mark(X))
p4: a__f#(X) -> mark#(X)
p5: mark#(f(X)) -> mark#(X)
p6: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p7: a__if#(false(),X,Y) -> mark#(Y)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(false(),X,Y) -> mark#(Y)
p3: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p4: a__if#(true(),X,Y) -> mark#(X)
p5: mark#(f(X)) -> mark#(X)
p6: mark#(f(X)) -> a__f#(mark(X))
p7: a__f#(X) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__f#_A(x1) = max{31, x1 + 11}
        a__if#_A(x1,x2,x3) = max{31, x1 - 4, x2 + 4, x3 + 10}
        mark_A(x1) = x1 + 10
        c_A = 0
        f_A(x1) = x1 + 20
        true_A = 0
        false_A = 36
        mark#_A(x1) = max{31, x1 + 4}
        if_A(x1,x2,x3) = max{28, x1 + 7, x2 + 10, x3 + 6}
        a__f_A(x1) = max{30, x1 + 20}
        a__if_A(x1,x2,x3) = max{29, x1 + 7, x2 + 10, x3 + 10}
  
    precedence:
    
      a__f# = a__if# = c = true = false = mark# > mark = f = if = a__f = a__if
    
    partial status:
  
      pi(a__f#) = []
      pi(a__if#) = []
      pi(mark) = []
      pi(c) = []
      pi(f) = []
      pi(true) = []
      pi(false) = []
      pi(mark#) = []
      pi(if) = []
      pi(a__f) = []
      pi(a__if) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)
p3: a__if#(true(),X,Y) -> mark#(X)
p4: mark#(f(X)) -> mark#(X)
p5: mark#(f(X)) -> a__f#(mark(X))
p6: a__f#(X) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: a__if#(true(),X,Y) -> mark#(X)
p3: mark#(f(X)) -> a__f#(mark(X))
p4: a__f#(X) -> mark#(X)
p5: mark#(f(X)) -> mark#(X)
p6: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__f#_A(x1) = max{5, x1 - 1}
        a__if#_A(x1,x2,x3) = max{x1 - 1, x2 + 3, x3 - 2}
        mark_A(x1) = x1
        c_A = 1
        f_A(x1) = max{7, x1}
        true_A = 6
        mark#_A(x1) = max{1, x1 - 1}
        if_A(x1,x2,x3) = max{x1, x2 + 5, x3}
        a__f_A(x1) = max{7, x1}
        a__if_A(x1,x2,x3) = max{x1, x2 + 5, x3}
        false_A = 1
  
    precedence:
    
      mark = true = if = a__f = a__if > c > a__f# = f = mark# > a__if# = false
    
    partial status:
  
      pi(a__f#) = []
      pi(a__if#) = []
      pi(mark) = []
      pi(c) = []
      pi(f) = []
      pi(true) = []
      pi(mark#) = []
      pi(if) = []
      pi(a__f) = []
      pi(a__if) = []
      pi(false) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(X) -> a__if#(mark(X),c(),f(true()))
p2: mark#(f(X)) -> a__f#(mark(X))
p3: a__f#(X) -> mark#(X)
p4: mark#(f(X)) -> mark#(X)
p5: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),mark(X2),X3)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(f(X)) -> a__f#(mark(X))
p2: a__f#(X) -> mark#(X)
p3: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{2, x1}
        f_A(x1) = max{26, x1 + 14}
        a__f#_A(x1) = x1 + 4
        mark_A(x1) = max{11, x1 + 9}
        a__f_A(x1) = max{35, x1 + 14}
        a__if_A(x1,x2,x3) = max{13, x1 + 5, x2 + 10, x3 + 9}
        c_A = 3
        true_A = 12
        false_A = 12
        if_A(x1,x2,x3) = max{12, x1 + 5, x2 + 10, x3 + 8}
  
    precedence:
    
      mark = a__f = a__if = if > f = a__f# > mark# > c = true = false
    
    partial status:
  
      pi(mark#) = [1]
      pi(f) = [1]
      pi(a__f#) = []
      pi(mark) = []
      pi(a__f) = []
      pi(a__if) = []
      pi(c) = []
      pi(true) = []
      pi(false) = []
      pi(if) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(f(X)) -> a__f#(mark(X))
p2: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(X) -> a__if(mark(X),c(),f(true()))
r2: a__if(true(),X,Y) -> mark(X)
r3: a__if(false(),X,Y) -> mark(Y)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(if(X1,X2,X3)) -> a__if(mark(X1),mark(X2),X3)
r6: mark(c()) -> c()
r7: mark(true()) -> true()
r8: mark(false()) -> false()
r9: a__f(X) -> f(X)
r10: a__if(X1,X2,X3) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        f_A(x1) = x1
  
    precedence:
    
      mark# = f
    
    partial status:
  
      pi(mark#) = [1]
      pi(f) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.