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:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{2, x2 + 1}
        c_A(x1) = max{14, x1 + 6}
        s_A(x1) = 0
        f_A(x1) = 22
        |0|_A = 22
        true_A = 21
        |1|_A = 0
        false_A = 5
        if_A(x1,x2,x3) = max{5, x1 - 29, x2 - 7, x3}
        g_A(x1,x2) = x2 + 7
        rand_A(x1) = x1 + 6
  
    precedence:
    
      g# = c = s = f = |0| = true = |1| = false = if = g = rand
    
    partial status:
  
      pi(g#) = [2]
      pi(c) = []
      pi(s) = []
      pi(f) = []
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []
      pi(if) = []
      pi(g) = []
      pi(rand) = []

The next rules are strictly ordered:

  p2

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#(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:

  {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)

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:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{28, x1 + 14, x2 + 18}
        c_A(x1) = x1 + 18
        s_A(x1) = 13
        f_A(x1) = 2
        |0|_A = 1
        true_A = 0
        |1|_A = 0
        false_A = 1
        if_A(x1,x2,x3) = x1 + 16
        g_A(x1,x2) = x2 + 26
        rand_A(x1) = max{28, x1 + 14}
  
    precedence:
    
      g# = c = s = f = |0| = true = |1| = false = if = g = rand
    
    partial status:
  
      pi(g#) = [2]
      pi(c) = []
      pi(s) = []
      pi(f) = []
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []
      pi(if) = [1]
      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:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1
        s_A(x1) = x1
        f_A(x1) = max{1, x1}
        |0|_A = 3
        true_A = 2
        |1|_A = 5
        false_A = 4
        if_A(x1,x2,x3) = max{7, x2 + 3, x3 + 3}
        g_A(x1,x2) = max{5, x2 - 2}
        c_A(x1) = 4
        rand_A(x1) = max{2, x1 + 1}
  
    precedence:
    
      f = |0| = true = |1| = false = if = g = c = rand > s > f#
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(|1|) = []
      pi(false) = []
      pi(if) = [3]
      pi(g) = []
      pi(c) = []
      pi(rand) = []

The next rules are strictly ordered:

  p1

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