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 path order with precedence:
  
    precedence:
    
      replace > if_replace > cons > s > min > nil > if_min > |sort|# > |0| > le > eq > false > true
    
    argument filter:
  
      pi(|sort|#) = [1]
      pi(cons) = [2]
      pi(replace) = 3
      pi(min) = [1]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []
      pi(false) = []
      pi(le) = []
      pi(if_min) = [2]
      pi(if_replace) = 4
      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 path order with precedence:
  
    precedence:
    
      cons > replace# > eq > s > true > |0| > false > if_replace#
    
    argument filter:
  
      pi(if_replace#) = [1, 4]
      pi(false) = []
      pi(cons) = [1, 2]
      pi(replace#) = [3]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []

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 monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > eq#
    
    argument filter:
  
      pi(eq#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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 path order with precedence:
  
    precedence:
    
      s > cons > |0| > le > true > false > min# > if_min#
    
    argument filter:
  
      pi(if_min#) = 2
      pi(false) = []
      pi(cons) = [2]
      pi(min#) = [1]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = []

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 monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > le#
    
    argument filter:
  
      pi(le#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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