YES

We show the termination of the TRS R:

  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(Y)) -> false()
  eq(s(X),|0|()) -> false()
  eq(s(X),s(Y)) -> eq(X,Y)
  le(|0|(),Y) -> true()
  le(s(X),|0|()) -> false()
  le(s(X),s(Y)) -> le(X,Y)
  min(cons(|0|(),nil())) -> |0|()
  min(cons(s(N),nil())) -> s(N)
  min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
  ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
  ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
  replace(N,M,nil()) -> nil()
  replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
  ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
  ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
  selsort(nil()) -> nil()
  selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
  ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
  ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

-- SCC decomposition.

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

p1: eq#(s(X),s(Y)) -> eq#(X,Y)
p2: le#(s(X),s(Y)) -> le#(X,Y)
p3: min#(cons(N,cons(M,L))) -> ifmin#(le(N,M),cons(N,cons(M,L)))
p4: min#(cons(N,cons(M,L))) -> le#(N,M)
p5: ifmin#(true(),cons(N,cons(M,L))) -> min#(cons(N,L))
p6: ifmin#(false(),cons(N,cons(M,L))) -> min#(cons(M,L))
p7: replace#(N,M,cons(K,L)) -> ifrepl#(eq(N,K),N,M,cons(K,L))
p8: replace#(N,M,cons(K,L)) -> eq#(N,K)
p9: ifrepl#(false(),N,M,cons(K,L)) -> replace#(N,M,L)
p10: selsort#(cons(N,L)) -> ifselsort#(eq(N,min(cons(N,L))),cons(N,L))
p11: selsort#(cons(N,L)) -> eq#(N,min(cons(N,L)))
p12: selsort#(cons(N,L)) -> min#(cons(N,L))
p13: ifselsort#(true(),cons(N,L)) -> selsort#(L)
p14: ifselsort#(false(),cons(N,L)) -> min#(cons(N,L))
p15: ifselsort#(false(),cons(N,L)) -> selsort#(replace(min(cons(N,L)),N,L))
p16: ifselsort#(false(),cons(N,L)) -> replace#(min(cons(N,L)),N,L)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

The estimated dependency graph contains the following SCCs:

  {p10, p13, p15}
  {p7, p9}
  {p1}
  {p3, p5, p6}
  {p2}


-- Reduction pair.

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

p1: ifselsort#(false(),cons(N,L)) -> selsort#(replace(min(cons(N,L)),N,L))
p2: selsort#(cons(N,L)) -> ifselsort#(eq(N,min(cons(N,L))),cons(N,L))
p3: ifselsort#(true(),cons(N,L)) -> selsort#(L)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

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. lexicographic path order with precedence:
    
      precedence:
      
        min > replace > nil > s > ifrepl > cons > |0| > ifmin > le > ifselsort# > selsort# > eq > true > false
      
      argument filter:
    
        pi(ifselsort#) = 2
        pi(false) = []
        pi(cons) = [2]
        pi(selsort#) = 1
        pi(replace) = 3
        pi(min) = 1
        pi(eq) = []
        pi(true) = []
        pi(le) = []
        pi(|0|) = []
        pi(s) = []
        pi(ifmin) = 2
        pi(ifrepl) = 4
        pi(nil) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > nil > false > eq > s > ifrepl > true > min > ifmin > le > replace > selsort# > cons > ifselsort#
      
      argument filter:
    
        pi(ifselsort#) = 2
        pi(false) = []
        pi(cons) = 2
        pi(selsort#) = 1
        pi(replace) = 3
        pi(min) = [1]
        pi(eq) = []
        pi(true) = []
        pi(le) = []
        pi(|0|) = []
        pi(s) = []
        pi(ifmin) = []
        pi(ifrepl) = 4
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: selsort#(cons(N,L)) -> ifselsort#(eq(N,min(cons(N,L))),cons(N,L))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: ifrepl#(false(),N,M,cons(K,L)) -> replace#(N,M,L)
p2: replace#(N,M,cons(K,L)) -> ifrepl#(eq(N,K),N,M,cons(K,L))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        eq > false > s > true > |0| > replace# > ifrepl# > cons
      
      argument filter:
    
        pi(ifrepl#) = 4
        pi(false) = []
        pi(cons) = [2]
        pi(replace#) = 3
        pi(eq) = [1, 2]
        pi(|0|) = []
        pi(true) = []
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        false > eq > s > true > |0| > ifrepl# > replace# > cons
      
      argument filter:
    
        pi(ifrepl#) = 4
        pi(false) = []
        pi(cons) = 2
        pi(replace#) = [3]
        pi(eq) = 2
        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(X),s(Y)) -> eq#(X,Y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > eq#
      
      argument filter:
    
        pi(eq#) = [2]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > eq#
      
      argument filter:
    
        pi(eq#) = [2]
        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: ifmin#(false(),cons(N,cons(M,L))) -> min#(cons(M,L))
p2: min#(cons(N,cons(M,L))) -> ifmin#(le(N,M),cons(N,cons(M,L)))
p3: ifmin#(true(),cons(N,cons(M,L))) -> min#(cons(N,L))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > false > |0| > true > cons > min# > le > ifmin#
      
      argument filter:
    
        pi(ifmin#) = [1, 2]
        pi(false) = []
        pi(cons) = [1, 2]
        pi(min#) = [1]
        pi(le) = [1]
        pi(true) = []
        pi(|0|) = []
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        false > le > s > true > |0| > cons > ifmin# > min#
      
      argument filter:
    
        pi(ifmin#) = 2
        pi(false) = []
        pi(cons) = 1
        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(X),s(Y)) -> le#(X,Y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(Y)) -> false()
r3: eq(s(X),|0|()) -> false()
r4: eq(s(X),s(Y)) -> eq(X,Y)
r5: le(|0|(),Y) -> true()
r6: le(s(X),|0|()) -> false()
r7: le(s(X),s(Y)) -> le(X,Y)
r8: min(cons(|0|(),nil())) -> |0|()
r9: min(cons(s(N),nil())) -> s(N)
r10: min(cons(N,cons(M,L))) -> ifmin(le(N,M),cons(N,cons(M,L)))
r11: ifmin(true(),cons(N,cons(M,L))) -> min(cons(N,L))
r12: ifmin(false(),cons(N,cons(M,L))) -> min(cons(M,L))
r13: replace(N,M,nil()) -> nil()
r14: replace(N,M,cons(K,L)) -> ifrepl(eq(N,K),N,M,cons(K,L))
r15: ifrepl(true(),N,M,cons(K,L)) -> cons(M,L)
r16: ifrepl(false(),N,M,cons(K,L)) -> cons(K,replace(N,M,L))
r17: selsort(nil()) -> nil()
r18: selsort(cons(N,L)) -> ifselsort(eq(N,min(cons(N,L))),cons(N,L))
r19: ifselsort(true(),cons(N,L)) -> cons(N,selsort(L))
r20: ifselsort(false(),cons(N,L)) -> cons(min(cons(N,L)),selsort(replace(min(cons(N,L)),N,L)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = [2]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > le#
      
      argument filter:
    
        pi(le#) = [2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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