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:

  max/plus interpretations on natural numbers:
  
    a__f#_A(x1) = max{22, x1 - 1}
    a__if#_A(x1,x2,x3) = max{8, x1 - 7, x2 + 1, x3 - 1}
    mark_A(x1) = x1 + 5
    c_A = 1
    f_A(x1) = max{23, x1 + 15}
    true_A = 8
    false_A = 8
    mark#_A(x1) = max{0, x1 - 1}
    if_A(x1,x2,x3) = max{x1 + 10, x2 + 8, x3 + 5}
    a__f_A(x1) = max{28, x1 + 15}
    a__if_A(x1,x2,x3) = max{x1 + 10, x2 + 8, x3 + 5}

The next rules are strictly ordered:

  p3, p4, p5, p6, p7

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#(f(X)) -> a__f#(mark(X))
p4: 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}


-- 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#(f(X)) -> a__f#(mark(X))
p4: 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:

  max/plus interpretations on natural numbers:
  
    a__f#_A(x1) = max{4, x1 + 2}
    a__if#_A(x1,x2,x3) = max{4, x1, x3 + 1}
    mark_A(x1) = x1 + 1
    c_A = 1
    f_A(x1) = max{3, x1 + 2}
    true_A = 1
    false_A = 1
    mark#_A(x1) = x1 + 1
    a__f_A(x1) = max{4, x1 + 2}
    a__if_A(x1,x2,x3) = max{2, x1 + 1, x2 + 1, x3 + 1}
    if_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3}

The next rules are strictly ordered:

  p4

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#(f(X)) -> a__f#(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}


-- 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#(f(X)) -> a__f#(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:

  max/plus interpretations on natural numbers:
  
    a__f#_A(x1) = max{5, x1 + 1}
    a__if#_A(x1,x2,x3) = max{0, x1 - 2, x3 - 1}
    mark_A(x1) = x1 + 2
    c_A = 2
    f_A(x1) = max{5, x1 + 4}
    true_A = 1
    false_A = 8
    mark#_A(x1) = max{5, x1 - 1}
    a__f_A(x1) = max{7, x1 + 4}
    a__if_A(x1,x2,x3) = max{x1 + 2, x2 + 2, x3 + 2}
    if_A(x1,x2,x3) = max{x1 + 2, x2 + 2, x3 + 2}

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: a__if#(false(),X,Y) -> mark#(Y)
p2: mark#(f(X)) -> a__f#(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:

  (no SCCs)