YES

We show the termination of the TRS R:

  active(f(f(X))) -> mark(c(f(g(f(X)))))
  active(c(X)) -> mark(d(X))
  active(h(X)) -> mark(c(d(X)))
  active(f(X)) -> f(active(X))
  active(h(X)) -> h(active(X))
  f(mark(X)) -> mark(f(X))
  h(mark(X)) -> mark(h(X))
  proper(f(X)) -> f(proper(X))
  proper(c(X)) -> c(proper(X))
  proper(g(X)) -> g(proper(X))
  proper(d(X)) -> d(proper(X))
  proper(h(X)) -> h(proper(X))
  f(ok(X)) -> ok(f(X))
  c(ok(X)) -> ok(c(X))
  g(ok(X)) -> ok(g(X))
  d(ok(X)) -> ok(d(X))
  h(ok(X)) -> ok(h(X))
  top(mark(X)) -> top(proper(X))
  top(ok(X)) -> top(active(X))

-- SCC decomposition.

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

p1: active#(f(f(X))) -> c#(f(g(f(X))))
p2: active#(f(f(X))) -> f#(g(f(X)))
p3: active#(f(f(X))) -> g#(f(X))
p4: active#(c(X)) -> d#(X)
p5: active#(h(X)) -> c#(d(X))
p6: active#(h(X)) -> d#(X)
p7: active#(f(X)) -> f#(active(X))
p8: active#(f(X)) -> active#(X)
p9: active#(h(X)) -> h#(active(X))
p10: active#(h(X)) -> active#(X)
p11: f#(mark(X)) -> f#(X)
p12: h#(mark(X)) -> h#(X)
p13: proper#(f(X)) -> f#(proper(X))
p14: proper#(f(X)) -> proper#(X)
p15: proper#(c(X)) -> c#(proper(X))
p16: proper#(c(X)) -> proper#(X)
p17: proper#(g(X)) -> g#(proper(X))
p18: proper#(g(X)) -> proper#(X)
p19: proper#(d(X)) -> d#(proper(X))
p20: proper#(d(X)) -> proper#(X)
p21: proper#(h(X)) -> h#(proper(X))
p22: proper#(h(X)) -> proper#(X)
p23: f#(ok(X)) -> f#(X)
p24: c#(ok(X)) -> c#(X)
p25: g#(ok(X)) -> g#(X)
p26: d#(ok(X)) -> d#(X)
p27: h#(ok(X)) -> h#(X)
p28: top#(mark(X)) -> top#(proper(X))
p29: top#(mark(X)) -> proper#(X)
p30: top#(ok(X)) -> top#(active(X))
p31: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p28, p30}
  {p8, p10}
  {p14, p16, p18, p20, p22}
  {p24}
  {p11, p23}
  {p25}
  {p26}
  {p12, p27}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))
p2: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          top#_A(x1) = ((1,0),(1,1)) x1 + (2,16)
          ok_A(x1) = ((1,0),(1,1)) x1 + (19,17)
          active_A(x1) = ((1,0),(1,1)) x1 + (14,1)
          mark_A(x1) = x1 + (2,17)
          proper_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          f_A(x1) = ((1,0),(1,1)) x1 + (15,16)
          h_A(x1) = ((1,0),(1,1)) x1 + (18,16)
          c_A(x1) = ((1,0),(1,1)) x1 + (9,7)
          g_A(x1) = ((1,0),(1,1)) x1 + (2,15)
          d_A(x1) = ((1,0),(1,1)) x1 + (8,2)
  
    precedence:
    
      top# = ok = active = mark = proper = f = h = c = g = d
    
    partial status:
  
      pi(top#) = [1]
      pi(ok) = [1]
      pi(active) = [1]
      pi(mark) = []
      pi(proper) = [1]
      pi(f) = [1]
      pi(h) = [1]
      pi(c) = [1]
      pi(g) = []
      pi(d) = [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: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          top#_A(x1) = x1 + (1,1)
          mark_A(x1) = ((1,0),(1,1)) x1 + (3,3)
          proper_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          f_A(x1) = ((1,0),(1,1)) x1 + (2,2)
          h_A(x1) = ((1,0),(1,1)) x1 + (2,2)
          ok_A(x1) = ((1,0),(1,1)) x1 + (3,3)
          c_A(x1) = ((1,0),(1,1)) x1 + (2,2)
          g_A(x1) = ((1,0),(1,1)) x1 + (2,2)
          d_A(x1) = ((1,0),(1,1)) x1 + (2,2)
  
    precedence:
    
      top# = mark = proper = f = h = ok = c = g = d
    
    partial status:
  
      pi(top#) = []
      pi(mark) = [1]
      pi(proper) = [1]
      pi(f) = [1]
      pi(h) = [1]
      pi(ok) = [1]
      pi(c) = [1]
      pi(g) = [1]
      pi(d) = [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#(h(X)) -> active#(X)
p2: active#(f(X)) -> active#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          active#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          h_A(x1) = x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      active# = h = f
    
    partial status:
  
      pi(active#) = []
      pi(h) = [1]
      pi(f) = []

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

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: active#(f(X)) -> active#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          active#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      active# = f
    
    partial status:
  
      pi(active#) = []
      pi(f) = []

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: proper#(h(X)) -> proper#(X)
p2: proper#(d(X)) -> proper#(X)
p3: proper#(g(X)) -> proper#(X)
p4: proper#(c(X)) -> proper#(X)
p5: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          proper#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          h_A(x1) = x1 + (3,1)
          d_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          g_A(x1) = ((1,0),(0,0)) x1 + (3,3)
          c_A(x1) = ((1,0),(1,1)) x1 + (3,3)
          f_A(x1) = ((1,0),(0,0)) x1 + (3,1)
  
    precedence:
    
      h = d = g > c = f > proper#
    
    partial status:
  
      pi(proper#) = []
      pi(h) = [1]
      pi(d) = []
      pi(g) = []
      pi(c) = []
      pi(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: proper#(h(X)) -> proper#(X)
p2: proper#(g(X)) -> proper#(X)
p3: proper#(c(X)) -> proper#(X)
p4: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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: proper#(h(X)) -> proper#(X)
p2: proper#(f(X)) -> proper#(X)
p3: proper#(c(X)) -> proper#(X)
p4: proper#(g(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          proper#_A(x1) = ((1,0),(0,0)) x1
          h_A(x1) = ((1,0),(1,0)) x1 + (1,1)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          c_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          g_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      proper# = h = f = g > c
    
    partial status:
  
      pi(proper#) = []
      pi(h) = []
      pi(f) = []
      pi(c) = []
      pi(g) = []

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: proper#(h(X)) -> proper#(X)
p2: proper#(f(X)) -> proper#(X)
p3: proper#(c(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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: proper#(h(X)) -> proper#(X)
p2: proper#(c(X)) -> proper#(X)
p3: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          proper#_A(x1) = ((1,0),(0,0)) x1
          h_A(x1) = ((1,0),(1,0)) x1 + (1,1)
          c_A(x1) = x1 + (1,1)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      proper# = h = f > c
    
    partial status:
  
      pi(proper#) = []
      pi(h) = []
      pi(c) = [1]
      pi(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: proper#(h(X)) -> proper#(X)
p2: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(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: proper#(h(X)) -> proper#(X)
p2: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          proper#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          h_A(x1) = x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      proper# = h = f
    
    partial status:
  
      pi(proper#) = []
      pi(h) = [1]
      pi(f) = []

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

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          proper#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      proper# = f
    
    partial status:
  
      pi(proper#) = []
      pi(f) = []

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: c#(ok(X)) -> c#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          c#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      c# = ok
    
    partial status:
  
      pi(c#) = []
      pi(ok) = []

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: f#(mark(X)) -> f#(X)
p2: f#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          f#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          mark_A(x1) = x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      f# = mark = ok
    
    partial status:
  
      pi(f#) = []
      pi(mark) = [1]
      pi(ok) = []

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#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(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#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          f#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      f# = ok
    
    partial status:
  
      pi(f#) = []
      pi(ok) = []

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: g#(ok(X)) -> g#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          g#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      g# = ok
    
    partial status:
  
      pi(g#) = []
      pi(ok) = []

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: d#(ok(X)) -> d#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          d#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      d# = ok
    
    partial status:
  
      pi(d#) = []
      pi(ok) = []

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: h#(mark(X)) -> h#(X)
p2: h#(ok(X)) -> h#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          h#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          mark_A(x1) = x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      h# = mark = ok
    
    partial status:
  
      pi(h#) = []
      pi(mark) = [1]
      pi(ok) = []

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: h#(ok(X)) -> h#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: h#(ok(X)) -> h#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

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:
          h#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      h# = ok
    
    partial status:
  
      pi(h#) = []
      pi(ok) = []

The next rules are strictly ordered:

  p1

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