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. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        |sort|#_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1
        cons_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + x2 + (2,0,0)
        replace_A(x1,x2,x3) = x2 + ((1,0,0),(1,1,0),(1,0,1)) x3 + (1,1,1)
        min_A(x1) = x1 + (2,6,1)
        eq_A(x1,x2) = ((1,0,0),(1,1,0),(0,1,0)) x1 + x2 + (1,1,0)
        |0|_A() = (3,0,1)
        true_A() = (0,3,1)
        s_A(x1) = ((1,0,0),(0,1,0),(1,0,1)) x1 + (1,2,1)
        false_A() = (0,3,1)
        le_A(x1,x2) = ((0,0,0),(1,0,0),(1,0,0)) x1 + ((0,0,0),(1,0,0),(1,0,0)) x2 + (1,1,1)
        if_min_A(x1,x2) = x1 + ((1,0,0),(1,1,0),(0,0,1)) x2
        if_replace_A(x1,x2,x3,x4) = x3 + ((1,0,0),(1,1,0),(1,0,1)) x4 + (1,0,0)
        nil_A() = (1,1,0)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > nil > if_replace > eq > replace > false > cons > min > le > if_min > true > |0| > |sort|#
      
      argument filter:
    
        pi(|sort|#) = []
        pi(cons) = 2
        pi(replace) = 3
        pi(min) = [1]
        pi(eq) = []
        pi(|0|) = []
        pi(true) = []
        pi(s) = 1
        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 combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        if_replace#_A(x1,x2,x3,x4) = x3 + ((1,0,0),(1,0,0),(0,1,0)) x4
        false_A() = (1,1,1)
        cons_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (2,2,1)
        replace#_A(x1,x2,x3) = x2 + ((1,0,0),(1,0,0),(0,1,0)) x3 + (1,1,1)
        eq_A(x1,x2) = x2 + (1,2,2)
        |0|_A() = (1,1,1)
        true_A() = (0,0,0)
        s_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > eq > cons > false > s > replace# > if_replace# > true
      
      argument filter:
    
        pi(if_replace#) = 3
        pi(false) = []
        pi(cons) = [2]
        pi(replace#) = 2
        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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        eq#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1
        s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > eq#
      
      argument filter:
    
        pi(eq#) = 1
        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. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        if_min#_A(x1,x2) = ((1,0,0),(1,1,0),(1,0,1)) x2
        false_A() = (1,1,6)
        cons_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (2,1,1)
        min#_A(x1) = ((1,0,0),(1,0,0),(1,0,0)) x1 + (1,0,0)
        le_A(x1,x2) = x1 + ((1,0,0),(0,1,0),(1,1,1)) x2 + (1,1,1)
        true_A() = (1,3,3)
        |0|_A() = (1,1,1)
        s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        le > false > s > true > |0| > cons > if_min# > min#
      
      argument filter:
    
        pi(if_min#) = []
        pi(false) = []
        pi(cons) = []
        pi(min#) = []
        pi(le) = [2]
        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. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        le#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1
        s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = 1
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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