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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifselsort#_A(x1,x2) = x1 + x2
        false_A() = 0
        cons_A(x1,x2) = x2 + 2
        selsort#_A(x1) = x1 + 2
        replace_A(x1,x2,x3) = x3
        min_A(x1) = x1 + 1
        eq_A(x1,x2) = 0
        true_A() = 0
        le_A(x1,x2) = 1
        |0|_A() = 1
        s_A(x1) = 1
        ifmin_A(x1,x2) = x2
        ifrepl_A(x1,x2,x3,x4) = x4
        nil_A() = 0
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifselsort#_A(x1,x2) = 1
        false_A() = 3
        cons_A(x1,x2) = 1
        selsort#_A(x1) = 0
        replace_A(x1,x2,x3) = 3
        min_A(x1) = 1
        eq_A(x1,x2) = 4
        true_A() = 1
        le_A(x1,x2) = 2
        |0|_A() = 2
        s_A(x1) = 2
        ifmin_A(x1,x2) = 0
        ifrepl_A(x1,x2,x3,x4) = 2
        nil_A() = 2
    

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: 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifrepl#_A(x1,x2,x3,x4) = x4
        false_A() = 1
        cons_A(x1,x2) = x1 + x2 + 2
        replace#_A(x1,x2,x3) = x3 + 1
        eq_A(x1,x2) = x2 + 1
        |0|_A() = 1
        true_A() = 0
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifrepl#_A(x1,x2,x3,x4) = x4
        false_A() = 3
        cons_A(x1,x2) = x1 + x2 + 2
        replace#_A(x1,x2,x3) = x3 + 1
        eq_A(x1,x2) = x2 + 1
        |0|_A() = 1
        true_A() = 0
        s_A(x1) = x1 + 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        eq#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        eq#_A(x1,x2) = 0
        s_A(x1) = x1 + 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifmin#_A(x1,x2) = x2
        false_A() = 1
        cons_A(x1,x2) = x1 + x2 + 2
        min#_A(x1) = x1 + 1
        le_A(x1,x2) = x1 + x2 + 1
        true_A() = 1
        |0|_A() = 1
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        ifmin#_A(x1,x2) = 0
        false_A() = 4
        cons_A(x1,x2) = x1 + 1
        min#_A(x1) = 1
        le_A(x1,x2) = x1 + x2 + 1
        true_A() = 3
        |0|_A() = 1
        s_A(x1) = x1 + 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(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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        le#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        le#_A(x1,x2) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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