YES

We show the termination of the TRS R:

  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(m)) -> false()
  eq(s(n),|0|()) -> false()
  eq(s(n),s(m)) -> eq(n,m)
  le(|0|(),m) -> true()
  le(s(n),|0|()) -> false()
  le(s(n),s(m)) -> le(n,m)
  min(cons(|0|(),nil())) -> |0|()
  min(cons(s(n),nil())) -> s(n)
  min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
  if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
  if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
  replace(n,m,nil()) -> nil()
  replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
  if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
  if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
  |sort|(nil()) -> nil()
  |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

-- SCC decomposition.

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

p1: eq#(s(n),s(m)) -> eq#(n,m)
p2: le#(s(n),s(m)) -> le#(n,m)
p3: min#(cons(n,cons(m,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))
p4: min#(cons(n,cons(m,x))) -> le#(n,m)
p5: if_min#(true(),cons(n,cons(m,x))) -> min#(cons(n,x))
p6: if_min#(false(),cons(n,cons(m,x))) -> min#(cons(m,x))
p7: replace#(n,m,cons(k,x)) -> if_replace#(eq(n,k),n,m,cons(k,x))
p8: replace#(n,m,cons(k,x)) -> eq#(n,k)
p9: if_replace#(false(),n,m,cons(k,x)) -> replace#(n,m,x)
p10: |sort|#(cons(n,x)) -> min#(cons(n,x))
p11: |sort|#(cons(n,x)) -> |sort|#(replace(min(cons(n,x)),n,x))
p12: |sort|#(cons(n,x)) -> replace#(min(cons(n,x)),n,x)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The estimated dependency graph contains the following SCCs:

  {p11}
  {p7, p9}
  {p1}
  {p3, p5, p6}
  {p2}


-- Reduction pair.

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

p1: |sort|#(cons(n,x)) -> |sort|#(replace(min(cons(n,x)),n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          |sort|#_A(x1) = max{6, x1 - 34}
          cons_A(x1,x2) = max{124, x2 + 41}
          replace_A(x1,x2,x3) = max{89, x3 + 8}
          min_A(x1) = max{186, x1 + 64}
          eq_A(x1,x2) = 132
          |0|_A = 189
          true_A = 125
          s_A(x1) = 128
          false_A = 125
          le_A(x1,x2) = 126
          if_min_A(x1,x2) = max{x1 + 102, x2 + 23}
          if_replace_A(x1,x2,x3,x4) = max{132, x1, x4 + 8}
          nil_A = 88
    
      precedence:
      
        true > |sort|# = cons = replace = eq = |0| = s = false = le = if_min = if_replace > min = nil
      
      partial status:
    
        pi(|sort|#) = []
        pi(cons) = []
        pi(replace) = [3]
        pi(min) = [1]
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(le) = []
        pi(if_min) = []
        pi(if_replace) = [4]
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          |sort|#_A(x1) = 16
          cons_A(x1,x2) = 18
          replace_A(x1,x2,x3) = x3 + 15
          min_A(x1) = 17
          eq_A(x1,x2) = 34
          |0|_A = 19
          true_A = 35
          s_A(x1) = 10
          false_A = 11
          le_A(x1,x2) = 19
          if_min_A(x1,x2) = 19
          if_replace_A(x1,x2,x3,x4) = x4 + 15
          nil_A = 15
    
      precedence:
      
        true > false = le = nil > |sort|# = cons = replace = eq = s = if_min = if_replace > |0| > min
      
      partial status:
    
        pi(|sort|#) = []
        pi(cons) = []
        pi(replace) = [3]
        pi(min) = []
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = []
        pi(false) = []
        pi(le) = []
        pi(if_min) = []
        pi(if_replace) = []
        pi(nil) = []
    

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: if_replace#(false(),n,m,cons(k,x)) -> replace#(n,m,x)
p2: replace#(n,m,cons(k,x)) -> if_replace#(eq(n,k),n,m,cons(k,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_replace#_A(x1,x2,x3,x4) = max{x2 + 9, x3 + 11, x4}
          false_A = 7
          cons_A(x1,x2) = max{12, x1 + 9, x2 + 7}
          replace#_A(x1,x2,x3) = max{x1 + 9, x2 + 11, x3}
          eq_A(x1,x2) = max{6, x1 + 1, x2 + 1}
          |0|_A = 8
          true_A = 7
          s_A(x1) = max{9, x1 + 8}
    
      precedence:
      
        if_replace# = replace# > |0| = true > cons = eq > false = s
      
      partial status:
    
        pi(if_replace#) = [2, 4]
        pi(false) = []
        pi(cons) = [2]
        pi(replace#) = [1, 3]
        pi(eq) = [2]
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_replace#_A(x1,x2,x3,x4) = max{x2 + 2, x4}
          false_A = 8
          cons_A(x1,x2) = max{3, x2 + 2}
          replace#_A(x1,x2,x3) = max{x1 + 2, x3}
          eq_A(x1,x2) = 4
          |0|_A = 9
          true_A = 10
          s_A(x1) = x1 + 9
    
      precedence:
      
        cons = replace# = true = s > if_replace# = false = eq = |0|
      
      partial status:
    
        pi(if_replace#) = [4]
        pi(false) = []
        pi(cons) = [2]
        pi(replace#) = [3]
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
    

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: eq#(s(n),s(m)) -> eq#(n,m)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

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:
        
          eq#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        eq# = s
      
      partial status:
    
        pi(eq#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          eq#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        eq# = s
      
      partial status:
    
        pi(eq#) = []
        pi(s) = [1]
    

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: if_min#(false(),cons(n,cons(m,x))) -> min#(cons(m,x))
p2: min#(cons(n,cons(m,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))
p3: if_min#(true(),cons(n,cons(m,x))) -> min#(cons(n,x))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_min#_A(x1,x2) = max{10, x1 + 1, x2}
          false_A = 11
          cons_A(x1,x2) = max{x1 + 6, x2 + 4}
          min#_A(x1) = max{7, x1}
          le_A(x1,x2) = x2 + 9
          true_A = 8
          |0|_A = 12
          s_A(x1) = x1 + 12
    
      precedence:
      
        cons = le > false > |0| > min# = true > if_min# = s
      
      partial status:
    
        pi(if_min#) = [1, 2]
        pi(false) = []
        pi(cons) = [1, 2]
        pi(min#) = [1]
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_min#_A(x1,x2) = max{x1 + 7, x2}
          false_A = 1
          cons_A(x1,x2) = max{x1 + 10, x2 + 4}
          min#_A(x1) = max{6, x1}
          le_A(x1,x2) = 0
          true_A = 1
          |0|_A = 0
          s_A(x1) = x1 + 2
    
      precedence:
      
        true > false = cons = le = |0| > if_min# = min# = s
      
      partial status:
    
        pi(if_min#) = [1, 2]
        pi(false) = []
        pi(cons) = [1, 2]
        pi(min#) = [1]
        pi(le) = []
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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: le#(s(n),s(m)) -> le#(n,m)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(m)) -> false()
r3: eq(s(n),|0|()) -> false()
r4: eq(s(n),s(m)) -> eq(n,m)
r5: le(|0|(),m) -> true()
r6: le(s(n),|0|()) -> false()
r7: le(s(n),s(m)) -> le(n,m)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(n),nil())) -> s(n)
r10: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r11: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r12: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r13: replace(n,m,nil()) -> nil()
r14: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r15: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r16: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r17: |sort|(nil()) -> nil()
r18: |sort|(cons(n,x)) -> cons(min(cons(n,x)),|sort|(replace(min(cons(n,x)),n,x)))

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:
        
          le#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        le# = s
      
      partial status:
    
        pi(le#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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