YES

We show the termination of the TRS R:

  a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
  mark(f(X1,X2)) -> a__f(mark(X1),X2)
  mark(g(X)) -> g(mark(X))
  a__f(X1,X2) -> f(X1,X2)

-- SCC decomposition.

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

p1: a__f#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: a__f#(g(X),Y) -> mark#(X)
p3: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)
p4: mark#(f(X1,X2)) -> mark#(X1)
p5: mark#(g(X)) -> mark#(X)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: a__f#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: a__f#(g(X),Y) -> mark#(X)
p3: mark#(g(X)) -> mark#(X)
p4: mark#(f(X1,X2)) -> mark#(X1)
p5: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1,x2) = ((1,0),(1,0)) x1 + (2,1)
          g_A(x1) = ((1,0),(0,0)) x1 + (5,4)
          mark_A(x1) = x1 + (1,7)
          f_A(x1,x2) = ((1,0),(1,0)) x1
          mark#_A(x1) = ((1,0),(1,0)) x1 + (4,3)
          a__f_A(x1,x2) = ((1,0),(1,0)) x1 + (0,4)
  
    precedence:
    
      a__f# = mark# > g = mark = f = a__f
    
    partial status:
  
      pi(a__f#) = []
      pi(g) = []
      pi(mark) = [1]
      pi(f) = []
      pi(mark#) = []
      pi(a__f) = []

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#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: a__f#(g(X),Y) -> mark#(X)
p3: mark#(f(X1,X2)) -> mark#(X1)
p4: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

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#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: a__f#(g(X),Y) -> mark#(X)
p3: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)
p4: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1,x2) = ((1,0),(0,0)) x1 + (3,2)
          g_A(x1) = x1 + (2,1)
          mark_A(x1) = ((1,0),(1,1)) x1 + (0,9)
          f_A(x1,x2) = ((1,0),(0,0)) x1 + (4,3)
          mark#_A(x1) = ((1,0),(1,0)) x1 + (1,4)
          a__f_A(x1,x2) = ((1,0),(0,0)) x1 + (4,4)
  
    precedence:
    
      a__f# = g = mark = f = mark# = a__f
    
    partial status:
  
      pi(a__f#) = []
      pi(g) = []
      pi(mark) = []
      pi(f) = []
      pi(mark#) = []
      pi(a__f) = []

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#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)
p3: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1}


-- Reduction pair.

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

p1: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          mark#_A(x1) = ((1,0),(1,1)) x1
          f_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
  
    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.

-- Reduction pair.

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

p1: a__f#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (1,5)
          g_A(x1) = ((1,0),(1,1)) x1 + (7,9)
          mark_A(x1) = ((1,0),(1,1)) x1 + (2,5)
          f_A(x1,x2) = x2 + (0,4)
          a__f_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,5)
  
    precedence:
    
      a__f# = g = mark = f = a__f
    
    partial status:
  
      pi(a__f#) = [1]
      pi(g) = []
      pi(mark) = [1]
      pi(f) = []
      pi(a__f) = []

The next rules are strictly ordered:

  p1

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