YES

We show the termination of the TRS R:

  f(|0|()) -> true()
  f(|1|()) -> false()
  f(s(x)) -> f(x)
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  g(s(x),s(y)) -> if(f(x),s(x),s(y))
  g(x,c(y)) -> g(x,g(s(c(y)),y))

-- SCC decomposition.

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

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

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

The estimated dependency graph contains the following SCCs:

  {p4, p5}
  {p1}


-- Reduction pair.

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

p1: g#(x,c(y)) -> g#(s(c(y)),y)
p2: g#(x,c(y)) -> g#(x,g(s(c(y)),y))

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

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,x2) = ((0,0),(1,0)) x2 + (7,2)
            c_A(x1) = x1 + (22,21)
            s_A(x1) = (6,9)
            g_A(x1,x2) = ((1,0),(1,0)) x1 + (14,20)
            f_A(x1) = (4,4)
            |0|_A() = (2,1)
            true_A() = (1,5)
            |1|_A() = (3,3)
            false_A() = (2,2)
            if_A(x1,x2,x3) = x1 + x2 + x3 + (2,2)
    
      precedence:
      
        g# = c = s = g = f = |0| = true = |1| = false = if
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(g) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = (0,0)
            c_A(x1) = x1 + (6,5)
            s_A(x1) = (2,1)
            g_A(x1,x2) = (5,4)
            f_A(x1) = (1,5)
            |0|_A() = (3,7)
            true_A() = (2,6)
            |1|_A() = (3,7)
            false_A() = (2,6)
            if_A(x1,x2,x3) = x2 + x3 + (1,1)
    
      precedence:
      
        if > f = true = |1| = false > |0| > g# = c = s = g
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
        pi(g) = []
        pi(f) = []
        pi(|0|) = []
        pi(true) = []
        pi(|1|) = []
        pi(false) = []
        pi(if) = []
    

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,c(y)) -> g#(s(c(y)),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

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,c(y)) -> g#(s(c(y)),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

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,x2) = x2
            c_A(x1) = ((1,0),(0,0)) x1 + (2,1)
            s_A(x1) = ((0,0),(1,0)) x1 + (1,1)
    
      precedence:
      
        c > g# = s
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1,x2) = (0,0)
            c_A(x1) = (2,2)
            s_A(x1) = (1,1)
    
      precedence:
      
        g# = c = s
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = []
    

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#(s(x)) -> f#(x)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),x,y) -> x
r5: if(false(),x,y) -> y
r6: g(s(x),s(y)) -> if(f(x),s(x),s(y))
r7: g(x,c(y)) -> g(x,g(s(c(y)),y))

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:
            f#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        f# = s
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = ((1,0),(0,0)) x1
            s_A(x1) = (1,1)
    
      precedence:
      
        s > f#
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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