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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          ifselsort#_A(x1,x2) = ((1,0),(0,0)) x2 + (4,15)
          false_A() = (0,4)
          cons_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (7,0)
          selsort#_A(x1) = x1 + (5,7)
          replace_A(x1,x2,x3) = ((1,0),(1,1)) x2 + x3 + (5,7)
          min_A(x1) = x1 + (6,0)
          eq_A(x1,x2) = x1 + x2 + (1,0)
          true_A() = (0,6)
          le_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (0,1)
          |0|_A() = (0,5)
          s_A(x1) = ((1,0),(1,1)) x1 + (1,2)
          ifmin_A(x1,x2) = ((1,0),(0,0)) x2 + (1,0)
          ifrepl_A(x1,x2,x3,x4) = ((1,0),(1,1)) x3 + x4 + (5,6)
          nil_A() = (2,1)
  
    precedence:
    
      ifselsort# = selsort# = replace = min = |0| = ifmin = ifrepl > cons > le = s > eq = true > false > nil
    
    partial status:
  
      pi(ifselsort#) = []
      pi(false) = []
      pi(cons) = [1]
      pi(selsort#) = []
      pi(replace) = [3]
      pi(min) = []
      pi(eq) = [1]
      pi(true) = []
      pi(le) = []
      pi(|0|) = []
      pi(s) = []
      pi(ifmin) = []
      pi(ifrepl) = [4]
      pi(nil) = []

The next rules are strictly ordered:

  p1

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))
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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))
p2: 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          selsort#_A(x1) = x1 + (3,4)
          cons_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (5,6)
          ifselsort#_A(x1,x2) = ((1,0),(0,0)) x2 + (0,5)
          eq_A(x1,x2) = (6,11)
          min_A(x1) = x1 + (4,5)
          true_A() = (1,3)
          le_A(x1,x2) = (2,2)
          |0|_A() = (0,0)
          s_A(x1) = (0,0)
          false_A() = (1,1)
          ifmin_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
          nil_A() = (1,1)
  
    precedence:
    
      selsort# = cons = ifselsort# = eq = min = true = le = |0| = s = false = ifmin = nil
    
    partial status:
  
      pi(selsort#) = []
      pi(cons) = []
      pi(ifselsort#) = []
      pi(eq) = []
      pi(min) = [1]
      pi(true) = []
      pi(le) = []
      pi(|0|) = []
      pi(s) = []
      pi(false) = []
      pi(ifmin) = [2]
      pi(nil) = []

The next rules are strictly ordered:

  p2

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          ifrepl#_A(x1,x2,x3,x4) = x4 + (1,1)
          false_A() = (1,4)
          cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (3,6)
          replace#_A(x1,x2,x3) = ((1,0),(1,1)) x3 + (2,5)
          eq_A(x1,x2) = ((1,0),(1,0)) x2 + (2,1)
          |0|_A() = (3,5)
          true_A() = (5,4)
          s_A(x1) = x1 + (3,5)
  
    precedence:
    
      cons > false > replace# > ifrepl# = |0| > eq = true > s
    
    partial status:
  
      pi(ifrepl#) = [4]
      pi(false) = []
      pi(cons) = [2]
      pi(replace#) = []
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          eq#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      s > eq#
    
    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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          ifmin#_A(x1,x2) = x2 + (4,1)
          false_A() = (3,2)
          cons_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (0,3)
          min#_A(x1) = x1 + (4,2)
          le_A(x1,x2) = ((1,0),(0,0)) x1 + (1,9)
          true_A() = (1,6)
          |0|_A() = (2,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (4,10)
  
    precedence:
    
      false = cons = le > ifmin# > min# = s > true = |0|
    
    partial status:
  
      pi(ifmin#) = []
      pi(false) = []
      pi(cons) = []
      pi(min#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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)))

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:

  {p1, p2}


-- 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)))

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          ifmin#_A(x1,x2) = x2 + (1,0)
          false_A() = (0,0)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (3,1)
          min#_A(x1) = ((1,0),(1,0)) x1 + (2,0)
          le_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,0)
          |0|_A() = (2,2)
          true_A() = (1,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      min# = le = s > |0| = true > cons > ifmin# = false
    
    partial status:
  
      pi(ifmin#) = [2]
      pi(false) = []
      pi(cons) = []
      pi(min#) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: ifmin#(false(),cons(N,cons(M,L))) -> min#(cons(M,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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          le#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      s > le#
    
    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.