YES
We show the termination of the TRS R:

  f(s(x)) -> f(id_inc(c(x,x)))
  f(c(s(x),y)) -> g(c(x,y))
  g(c(s(x),y)) -> g(c(y,x))
  g(c(x,s(y))) -> g(c(y,x))
  g(c(x,x)) -> f(x)
  id_inc(c(x,y)) -> c(id_inc(x),id_inc(y))
  id_inc(s(x)) -> s(id_inc(x))
  id_inc(|0|()) -> |0|()
  id_inc(|0|()) -> s(|0|())


-- SCC decomposition.

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

p1: f#(s(x)) -> f#(id_inc(c(x,x)))
p2: f#(s(x)) -> id_inc#(c(x,x))
p3: f#(c(s(x),y)) -> g#(c(x,y))
p4: g#(c(s(x),y)) -> g#(c(y,x))
p5: g#(c(x,s(y))) -> g#(c(y,x))
p6: g#(c(x,x)) -> f#(x)
p7: id_inc#(c(x,y)) -> id_inc#(x)
p8: id_inc#(c(x,y)) -> id_inc#(y)
p9: id_inc#(s(x)) -> id_inc#(x)

and R consists of:

r1: f(s(x)) -> f(id_inc(c(x,x)))
r2: f(c(s(x),y)) -> g(c(x,y))
r3: g(c(s(x),y)) -> g(c(y,x))
r4: g(c(x,s(y))) -> g(c(y,x))
r5: g(c(x,x)) -> f(x)
r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y))
r7: id_inc(s(x)) -> s(id_inc(x))
r8: id_inc(|0|()) -> |0|()
r9: id_inc(|0|()) -> s(|0|())

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4, p5, p6}
  {p7, p8, p9}


-- Reduction pair.

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

p1: f#(s(x)) -> f#(id_inc(c(x,x)))
p2: f#(c(s(x),y)) -> g#(c(x,y))
p3: g#(c(x,x)) -> f#(x)
p4: g#(c(x,s(y))) -> g#(c(y,x))
p5: g#(c(s(x),y)) -> g#(c(y,x))

and R consists of:

r1: f(s(x)) -> f(id_inc(c(x,x)))
r2: f(c(s(x),y)) -> g(c(x,y))
r3: g(c(s(x),y)) -> g(c(y,x))
r4: g(c(x,s(y))) -> g(c(y,x))
r5: g(c(x,x)) -> f(x)
r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y))
r7: id_inc(s(x)) -> s(id_inc(x))
r8: id_inc(|0|()) -> |0|()
r9: id_inc(|0|()) -> s(|0|())

The set of usable rules consists of

  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!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1) = 1
            s_A(x1) = max{30, x1 + 14}
            id_inc_A(x1) = x1 + 14
            c_A(x1,x2) = max{15, x1 - 4, x2}
            g#_A(x1) = 1
            |0|_A = 16
            f#!_A(x1) = max{21, x1 + 8}
            s!_A(x1) = 75
            id_inc!_A(x1) = x1 + 44
            c!_A(x1,x2) = max{x1 + 8, x2 + 3}
            g#!_A(x1) = x1 + 8
            |0|!_A = 75
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = 0
              s_A(x1) = x1 + 4
              id_inc_A(x1) = x1 + 4
              c_A(x1,x2) = x1
              g#_A(x1) = 0
              |0|_A() = 4
              f#!_A(x1) = x1 + 8
              s!_A(x1) = 7
              id_inc!_A(x1) = x1 + 4
              c!_A(x1,x2) = x2 + 6
              g#!_A(x1) = x1 + 8
              |0|!_A() = 7
        
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(id_inc) = [1]
      pi(c) = []
      pi(g#) = [1]
      pi(|0|) = []

The following rules are removed:

  p2, p3, p4, p5



-- SCC decomposition.

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

p1: f#(s(x)) -> f#(id_inc(c(x,x)))

and R consists of:

r1: f(s(x)) -> f(id_inc(c(x,x)))
r2: f(c(s(x),y)) -> g(c(x,y))
r3: g(c(s(x),y)) -> g(c(y,x))
r4: g(c(x,s(y))) -> g(c(y,x))
r5: g(c(x,x)) -> f(x)
r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y))
r7: id_inc(s(x)) -> s(id_inc(x))
r8: id_inc(|0|()) -> |0|()
r9: id_inc(|0|()) -> s(|0|())

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(s(x)) -> f#(id_inc(c(x,x)))

and R consists of:

r1: f(s(x)) -> f(id_inc(c(x,x)))
r2: f(c(s(x),y)) -> g(c(x,y))
r3: g(c(s(x),y)) -> g(c(y,x))
r4: g(c(x,s(y))) -> g(c(y,x))
r5: g(c(x,x)) -> f(x)
r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y))
r7: id_inc(s(x)) -> s(id_inc(x))
r8: id_inc(|0|()) -> |0|()
r9: id_inc(|0|()) -> s(|0|())

The set of usable rules consists of

  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!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1) = 0
            s_A(x1) = max{0, x1 - 6}
            id_inc_A(x1) = max{0, x1 - 12}
            c_A(x1,x2) = max{0, x1 - 12, x2 - 6}
            |0|_A = 0
            f#!_A(x1) = 10
            s!_A(x1) = x1 + 11
            id_inc!_A(x1) = 9
            c!_A(x1,x2) = max{8, x1 + 7}
            |0|!_A = 12
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = 0
              s_A(x1) = x1
              id_inc_A(x1) = 1
              c_A(x1,x2) = x2
              |0|_A() = 1
              f#!_A(x1) = 5
              s!_A(x1) = 4
              id_inc!_A(x1) = 0
              c!_A(x1,x2) = 3
              |0|!_A() = 5
        
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = []
      pi(id_inc) = [1]
      pi(c) = []
      pi(|0|) = []

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: id_inc#(c(x,y)) -> id_inc#(x)
p2: id_inc#(s(x)) -> id_inc#(x)
p3: id_inc#(c(x,y)) -> id_inc#(y)

and R consists of:

r1: f(s(x)) -> f(id_inc(c(x,x)))
r2: f(c(s(x),y)) -> g(c(x,y))
r3: g(c(s(x),y)) -> g(c(y,x))
r4: g(c(x,s(y))) -> g(c(y,x))
r5: g(c(x,x)) -> f(x)
r6: id_inc(c(x,y)) -> c(id_inc(x),id_inc(y))
r7: id_inc(s(x)) -> s(id_inc(x))
r8: id_inc(|0|()) -> |0|()
r9: id_inc(|0|()) -> s(|0|())

The set of usable rules consists of

  (no rules)

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!):
        
          max/plus interpretation on natural numbers:
          
            id_inc#_A(x1) = 0
            c_A(x1,x2) = max{2, x1 - 9, x2 - 10}
            s_A(x1) = max{6, x1 - 10}
            id_inc#!_A(x1) = 3
            c!_A(x1,x2) = max{2, x1 - 19, x2 - 24}
            s!_A(x1) = max{0, x1 - 22}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              id_inc#_A(x1) = x1
              c_A(x1,x2) = x1 + x2 + 1
              s_A(x1) = x1 + 1
              id_inc#!_A(x1) = x1 + 6
              c!_A(x1,x2) = 5
              s!_A(x1) = x1
        
  
    partial status:
  
      pi(id_inc#) = [1]
      pi(c) = [1, 2]
      pi(s) = [1]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.