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, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              tp#_A(x1) = 0
              mark_A(x1) = 5
              chk_A(x1) = x1
              mat_A(x1,x2) = x2
              f_A(x1) = 7
              X_A() = 7
              active_A(x1) = 5
              no_A(x1) = x1
              y_A() = 1
              c_A() = 15
              tp_A(x1) = 0
              tp#!_A(x1) = 0
              mark!_A(x1) = x1 + 7
              chk!_A(x1) = x1 + 1
              mat!_A(x1,x2) = x2 + 6
              f!_A(x1) = x1 + 10
              X!_A() = 8
              active!_A(x1) = 15
              no!_A(x1) = 9
              y!_A() = 12
              c!_A() = 17
              tp!_A(x1) = 17
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              tp#_A(x1) = 0
              mark_A(x1) = 0
              chk_A(x1) = 0
              mat_A(x1,x2) = 2
              f_A(x1) = x1
              X_A() = 3
              active_A(x1) = 0
              no_A(x1) = x1 + 1
              y_A() = 3
              c_A() = 1
              tp_A(x1) = 0
              tp#!_A(x1) = 2
              mark!_A(x1) = 6
              chk!_A(x1) = x1 + 2
              mat!_A(x1,x2) = 1
              f!_A(x1) = x1 + 7
              X!_A() = 3
              active!_A(x1) = 5
              no!_A(x1) = x1 + 9
              y!_A() = 0
              c!_A() = 10
              tp!_A(x1) = 11
        
  
    partial status:
  
      pi(tp#) = [1]
      pi(mark) = []
      pi(chk) = [1]
      pi(mat) = []
      pi(f) = []
      pi(X) = []
      pi(active) = []
      pi(no) = [1]
      pi(y) = []
      pi(c) = []
      pi(tp) = [1]

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              chk#_A(x1) = x1
              no_A(x1) = x1 + 1
              f_A(x1) = x1 + 2
              mat_A(x1,x2) = x2 + 2
              X_A() = 1
              y_A() = 1
              c_A() = 1
              chk#!_A(x1) = x1 + 2
              no!_A(x1) = 3
              f!_A(x1) = 1
              mat!_A(x1,x2) = x2 + 6
              X!_A() = 0
              y!_A() = 2
              c!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              chk#_A(x1) = x1
              no_A(x1) = x1 + 1
              f_A(x1) = x1 + 2
              mat_A(x1,x2) = x2 + 2
              X_A() = 1
              y_A() = 1
              c_A() = 1
              chk#!_A(x1) = 3
              no!_A(x1) = 4
              f!_A(x1) = 2
              mat!_A(x1,x2) = 1
              X!_A() = 1
              y!_A() = 3
              c!_A() = 0
        
  
    partial status:
  
      pi(chk#) = []
      pi(no) = [1]
      pi(f) = []
      pi(mat) = [1, 2]
      pi(X) = []
      pi(y) = []
      pi(c) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              active#_A(x1) = 0
              f_A(x1) = x1 + 2
              f#_A(x1) = 0
              mark_A(x1) = x1 + 24
              no_A(x1) = 28
              active_A(x1) = x1 + 53
              active#!_A(x1) = 0
              f!_A(x1) = x1 + 1
              f#!_A(x1) = 0
              mark!_A(x1) = x1 + 22
              no!_A(x1) = 28
              active!_A(x1) = x1 + 25
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              active#_A(x1) = 0
              f_A(x1) = x1 + 3
              f#_A(x1) = 0
              mark_A(x1) = x1 + 36
              no_A(x1) = x1 + 29
              active_A(x1) = x1 + 40
              active#!_A(x1) = x1 + 1
              f!_A(x1) = x1 + 5
              f#!_A(x1) = x1
              mark!_A(x1) = x1 + 37
              no!_A(x1) = 33
              active!_A(x1) = x1 + 41
        
  
    partial status:
  
      pi(active#) = []
      pi(f) = []
      pi(f#) = []
      pi(mark) = []
      pi(no) = []
      pi(active) = []

The following rules are removed:

  p1, p2, p3, p4, p5

Then no dependency pair remains.