YES

We show the termination of the TRS R:

  f(nil()) -> nil()
  f(.(nil(),y)) -> .(nil(),f(y))
  f(.(.(x,y),z)) -> f(.(x,.(y,z)))
  g(nil()) -> nil()
  g(.(x,nil())) -> .(g(x),nil())
  g(.(x,.(y,z))) -> g(.(.(x,y),z))

-- SCC decomposition.

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

p1: f#(.(nil(),y)) -> f#(y)
p2: f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))
p3: g#(.(x,nil())) -> g#(x)
p4: g#(.(x,.(y,z))) -> g#(.(.(x,y),z))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

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: f#(.(nil(),y)) -> f#(y)
p2: f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 1
          ._A(x1,x2) = max{x1, x2}
          nil_A = 0
    
      precedence:
      
        f# = . = nil
      
      partial status:
    
        pi(f#) = [1]
        pi(.) = [1, 2]
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 6
          ._A(x1,x2) = max{x1 + 2, x2}
          nil_A = 0
    
      precedence:
      
        f# = . = nil
      
      partial status:
    
        pi(f#) = [1]
        pi(.) = [1, 2]
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p2

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#(.(x,nil())) -> g#(x)
p2: g#(.(x,.(y,z))) -> g#(.(.(x,y),z))

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = x1 + (0,2)
            ._A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,3)
            nil_A() = (1,1)
    
      precedence:
      
        g# = . = nil
      
      partial status:
    
        pi(g#) = []
        pi(.) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = (0,0)
            ._A(x1,x2) = ((1,0),(0,0)) x2 + (1,1)
            nil_A() = (1,1)
    
      precedence:
      
        g# = . > nil
      
      partial status:
    
        pi(g#) = []
        pi(.) = []
        pi(nil) = []
    

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: g#(.(x,nil())) -> g#(x)

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: g#(.(x,nil())) -> g#(x)

and R consists of:

r1: f(nil()) -> nil()
r2: f(.(nil(),y)) -> .(nil(),f(y))
r3: f(.(.(x,y),z)) -> f(.(x,.(y,z)))
r4: g(nil()) -> nil()
r5: g(.(x,nil())) -> .(g(x),nil())
r6: g(.(x,.(y,z))) -> g(.(.(x,y),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{2, x1 + 1}
          ._A(x1,x2) = max{x1, x2}
          nil_A = 0
    
      precedence:
      
        g# = . = nil
      
      partial status:
    
        pi(g#) = [1]
        pi(.) = [1, 2]
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{1, x1 - 1}
          ._A(x1,x2) = x2
          nil_A = 0
    
      precedence:
      
        g# = . = nil
      
      partial status:
    
        pi(g#) = []
        pi(.) = [2]
        pi(nil) = []
    

The next rules are strictly ordered:

  p1

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