YES

We show the termination of the relative TRS R/S:

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

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

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

p1: f#(s(x)) -> f#(x)
p2: g#(x,c(y)) -> g#(x,y)
p3: g#(x,c(y)) -> g#(x,if(f(x),c(g(s(x),y)),c(y)))
p4: g#(x,c(y)) -> if#(f(x),c(g(s(x),y)),c(y))
p5: g#(x,c(y)) -> f#(x)
p6: g#(x,c(y)) -> g#(s(x),y)

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p2, p6}
  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: f(|0|()) -> true()
r2: f(|1|()) -> false()
r3: f(s(x)) -> f(x)
r4: if(true(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

Consider the non-minimal 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(),s(x),s(y)) -> s(x)
r5: if(false(),s(x),s(y)) -> s(y)
r6: g(x,c(y)) -> c(g(x,y))
r7: g(x,c(y)) -> g(x,if(f(x),c(g(s(x),y)),c(y)))
r8: rand(x) -> x
r9: rand(x) -> rand(s(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = x1 + (0,3)
          s_A(x1) = x1 + (0,4)
          f_A(x1) = (2,6)
          |0|_A() = (2,6)
          true_A() = (1,5)
          |1|_A() = (2,1)
          false_A() = (1,7)
          if_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (1,2)
          g_A(x1,x2) = (8,1)
          c_A(x1) = (3,3)
          rand_A(x1) = ((1,0),(0,0)) x1 + (1,5)
  
    precedence:
    
      true = |1| = false = rand > f# > |0| > f = g = c > s = if
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = []
      pi(f) = []
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []
      pi(if) = []
      pi(g) = []
      pi(c) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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