YES

We show the termination of the TRS R:

  active(f(x)) -> mark(f(f(x)))
  chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
  mat(f(x),f(y())) -> f(mat(x,y()))
  chk(no(c())) -> active(c())
  mat(f(x),c()) -> no(c())
  f(active(x)) -> active(f(x))
  f(no(x)) -> no(f(x))
  f(mark(x)) -> mark(f(x))
  tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

-- SCC decomposition.

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

p1: active#(f(x)) -> f#(f(x))
p2: chk#(no(f(x))) -> f#(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
p3: chk#(no(f(x))) -> chk#(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x))
p4: chk#(no(f(x))) -> mat#(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)
p5: chk#(no(f(x))) -> f#(f(f(f(f(f(f(f(f(f(X()))))))))))
p6: chk#(no(f(x))) -> f#(f(f(f(f(f(f(f(f(X())))))))))
p7: chk#(no(f(x))) -> f#(f(f(f(f(f(f(f(X()))))))))
p8: chk#(no(f(x))) -> f#(f(f(f(f(f(f(X())))))))
p9: chk#(no(f(x))) -> f#(f(f(f(f(f(X()))))))
p10: chk#(no(f(x))) -> f#(f(f(f(f(X())))))
p11: chk#(no(f(x))) -> f#(f(f(f(X()))))
p12: chk#(no(f(x))) -> f#(f(f(X())))
p13: chk#(no(f(x))) -> f#(f(X()))
p14: chk#(no(f(x))) -> f#(X())
p15: mat#(f(x),f(y())) -> f#(mat(x,y()))
p16: mat#(f(x),f(y())) -> mat#(x,y())
p17: chk#(no(c())) -> active#(c())
p18: f#(active(x)) -> active#(f(x))
p19: f#(active(x)) -> f#(x)
p20: f#(no(x)) -> f#(x)
p21: f#(mark(x)) -> f#(x)
p22: tp#(mark(x)) -> tp#(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
p23: tp#(mark(x)) -> chk#(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x))
p24: tp#(mark(x)) -> mat#(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)
p25: tp#(mark(x)) -> f#(f(f(f(f(f(f(f(f(f(X()))))))))))
p26: tp#(mark(x)) -> f#(f(f(f(f(f(f(f(f(X())))))))))
p27: tp#(mark(x)) -> f#(f(f(f(f(f(f(f(X()))))))))
p28: tp#(mark(x)) -> f#(f(f(f(f(f(f(X())))))))
p29: tp#(mark(x)) -> f#(f(f(f(f(f(X()))))))
p30: tp#(mark(x)) -> f#(f(f(f(f(X())))))
p31: tp#(mark(x)) -> f#(f(f(f(X()))))
p32: tp#(mark(x)) -> f#(f(f(X())))
p33: tp#(mark(x)) -> f#(f(X()))
p34: tp#(mark(x)) -> f#(X())

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

The estimated dependency graph contains the following SCCs:

  {p22}
  {p3}
  {p1, p18, p19, p20, p21}


-- Reduction pair.

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

p1: tp#(mark(x)) -> tp#(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        tp#_A(x1) = max{111, x1 + 105}
        mark_A(x1) = max{34, x1 - 2}
        chk_A(x1) = max{19, x1 - 132}
        mat_A(x1,x2) = max{153, x2 + 24}
        f_A(x1) = x1 + 23
        X_A = 1
        active_A(x1) = x1 + 21
        no_A(x1) = x1 + 153
        y_A = 154
        c_A = 0
  
    precedence:
    
      mat > mark = chk = f = X = active = no = c > tp# = y
    
    partial status:
  
      pi(tp#) = [1]
      pi(mark) = []
      pi(chk) = []
      pi(mat) = []
      pi(f) = []
      pi(X) = []
      pi(active) = []
      pi(no) = []
      pi(y) = []
      pi(c) = []

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: chk#(no(f(x))) -> chk#(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x))

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

The set of usable rules consists of

  r3, r5

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        chk#_A(x1) = max{104, x1 + 65}
        no_A(x1) = x1 + 70
        f_A(x1) = max{33, x1 + 18}
        mat_A(x1,x2) = x2 + 71
        X_A = 17
        y_A = 1
        c_A = 17
  
    precedence:
    
      chk# = no = f = mat = X = y = c
    
    partial status:
  
      pi(chk#) = [1]
      pi(no) = []
      pi(f) = [1]
      pi(mat) = [2]
      pi(X) = []
      pi(y) = []
      pi(c) = []

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: active#(f(x)) -> f#(f(x))
p2: f#(mark(x)) -> f#(x)
p3: f#(no(x)) -> f#(x)
p4: f#(active(x)) -> f#(x)
p5: f#(active(x)) -> active#(f(x))

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

The set of usable rules consists of

  r1, r6, r7, r8

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1 + 10
        f_A(x1) = x1 + 2
        f#_A(x1) = max{12, x1 + 7}
        mark_A(x1) = x1 + 3
        no_A(x1) = x1 + 6
        active_A(x1) = x1 + 6
  
    precedence:
    
      active# = f = f# = mark = no = active
    
    partial status:
  
      pi(active#) = [1]
      pi(f) = [1]
      pi(f#) = []
      pi(mark) = [1]
      pi(no) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(f(x)) -> f#(f(x))
p2: f#(mark(x)) -> f#(x)
p3: f#(no(x)) -> f#(x)
p4: f#(active(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

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: f#(mark(x)) -> f#(x)
p2: f#(active(x)) -> f#(x)
p3: f#(no(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

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:
      
        f#_A(x1) = x1 + 2
        mark_A(x1) = x1 + 1
        active_A(x1) = x1
        no_A(x1) = x1
  
    precedence:
    
      f# = mark = active = no
    
    partial status:
  
      pi(f#) = [1]
      pi(mark) = [1]
      pi(active) = [1]
      pi(no) = [1]

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: f#(active(x)) -> f#(x)
p2: f#(no(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(active(x)) -> f#(x)
p2: f#(no(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

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:
      
        f#_A(x1) = x1 + 1
        active_A(x1) = x1
        no_A(x1) = x1
  
    precedence:
    
      f# = active = no
    
    partial status:
  
      pi(f#) = [1]
      pi(active) = [1]
      pi(no) = [1]

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: f#(no(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(no(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(f(f(x)))
r2: chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
r3: mat(f(x),f(y())) -> f(mat(x,y()))
r4: chk(no(c())) -> active(c())
r5: mat(f(x),c()) -> no(c())
r6: f(active(x)) -> active(f(x))
r7: f(no(x)) -> no(f(x))
r8: f(mark(x)) -> mark(f(x))
r9: tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))

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:
      
        f#_A(x1) = x1 + 2
        no_A(x1) = x1 + 2
  
    precedence:
    
      f# = no
    
    partial status:
  
      pi(f#) = [1]
      pi(no) = [1]

The next rules are strictly ordered:

  p1

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