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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      tp#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,0,1)) x1
      mark_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + (0,1,6,6)
      chk_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + (0,1,5,4)
      mat_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + (0,0,0,3)
      f_A(x1) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + (0,1,6,8)
      X_A() = (1,1,0,0)
      active_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,0)) x1 + (0,0,0,1)
      no_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (0,0,4,7)
      y_A() = (5,1,1,1)
      c_A() = (0,0,1,0)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      chk#_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,1,0,0)) x1
      no_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (0,0,0,3)
      f_A(x1) = x1 + (3,1,2,1)
      mat_A(x1,x2) = x2 + (1,2,1,2)
      X_A() = (1,1,1,0)
      y_A() = (1,1,1,1)
      c_A() = (1,3,1,1)

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: 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,0,1,1)) x1 + (1,6,0,0)
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,1,6,11)
      f#_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,1)) x1 + (0,0,1,8)
      mark_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,0,1)) x1 + (2,4,1,1)
      no_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,0,0),(0,0,0,0)) x1 + (1,1,1,15)
      active_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,1,0,1)) x1 + (4,6,1,9)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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