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:
  
      max/plus interpretations on natural numbers:
      
        ifselsort#_A(x1,x2) = max{x1 - 2, x2 + 46}
        false_A = 97
        cons_A(x1,x2) = max{99, x2 + 55}
        selsort#_A(x1) = x1 + 91
        replace_A(x1,x2,x3) = x3 + 10
        min_A(x1) = max{94, x1 + 18}
        eq_A(x1,x2) = 100
        true_A = 97
        le_A(x1,x2) = 98
        |0|_A = 100
        s_A(x1) = 178
        ifmin_A(x1,x2) = max{93, x1 + 11, x2 - 16}
        ifrepl_A(x1,x2,x3,x4) = max{59, x1 + 9, x4 + 10}
        nil_A = 124
  
    precedence:
    
      ifselsort# = false = selsort# = replace = min = eq = true = le = |0| = s = ifmin > cons = ifrepl > nil
    
    partial status:
  
      pi(ifselsort#) = [2]
      pi(false) = []
      pi(cons) = []
      pi(selsort#) = [1]
      pi(replace) = [3]
      pi(min) = []
      pi(eq) = []
      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:
  
      max/plus interpretations on natural numbers:
      
        selsort#_A(x1) = max{24, x1 + 16}
        cons_A(x1,x2) = max{x1 + 20, x2 + 6}
        ifselsort#_A(x1,x2) = max{21, x1 - 32, x2 + 10}
        eq_A(x1,x2) = max{37, x1 + 19}
        min_A(x1) = max{37, x1 + 27}
        true_A = 57
        le_A(x1,x2) = max{x1 + 63, x2 + 69}
        |0|_A = 39
        s_A(x1) = max{94, x1 + 1}
        false_A = 58
        ifmin_A(x1,x2) = max{38, x1 - 16, x2 + 27}
        nil_A = 34
  
    precedence:
    
      selsort# = cons = ifselsort# = eq = min = true = le = |0| = s = false = ifmin = nil
    
    partial status:
  
      pi(selsort#) = []
      pi(cons) = [1, 2]
      pi(ifselsort#) = [2]
      pi(eq) = []
      pi(min) = [1]
      pi(true) = []
      pi(le) = [2]
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []
      pi(ifmin) = [2]
      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: 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:

  (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:
  
      max/plus interpretations on natural numbers:
      
        ifrepl#_A(x1,x2,x3,x4) = max{3, x1 - 1, x2 + 2, x3 + 2, x4}
        false_A = 3
        cons_A(x1,x2) = max{3, x1, x2 + 2}
        replace#_A(x1,x2,x3) = max{x1 + 2, x2 + 2, x3}
        eq_A(x1,x2) = 4
        |0|_A = 1
        true_A = 0
        s_A(x1) = max{10, x1 + 4}
  
    precedence:
    
      eq > ifrepl# = false = cons = replace# = |0| = true = s
    
    partial status:
  
      pi(ifrepl#) = [4]
      pi(false) = []
      pi(cons) = [1]
      pi(replace#) = [3]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]

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: 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 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:
  
      max/plus interpretations on natural numbers:
      
        eq#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      eq# = s
    
    partial status:
  
      pi(eq#) = []
      pi(s) = []

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:
  
      max/plus interpretations on natural numbers:
      
        ifmin#_A(x1,x2) = max{31, x1 - 16, x2 + 4}
        false_A = 35
        cons_A(x1,x2) = max{18, x2 + 12}
        min#_A(x1) = max{21, x1 + 5}
        le_A(x1,x2) = 36
        true_A = 35
        |0|_A = 0
        s_A(x1) = x1 + 36
  
    precedence:
    
      ifmin# = false = cons = min# = le = true = |0| = s
    
    partial status:
  
      pi(ifmin#) = [2]
      pi(false) = []
      pi(cons) = [2]
      pi(min#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

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: min#(cons(N,cons(M,L))) -> ifmin#(le(N,M),cons(N,cons(M,L)))
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: min#(cons(N,cons(M,L))) -> ifmin#(le(N,M),cons(N,cons(M,L)))
p2: 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:
  
      max/plus interpretations on natural numbers:
      
        min#_A(x1) = x1 + 9
        cons_A(x1,x2) = max{5, x2 + 4}
        ifmin#_A(x1,x2) = max{10, x1 - 3, x2 + 5}
        le_A(x1,x2) = 10
        true_A = 9
        |0|_A = 0
        s_A(x1) = x1 + 2
        false_A = 1
  
    precedence:
    
      min# = cons = ifmin# = le = true = |0| = s = false
    
    partial status:
  
      pi(min#) = []
      pi(cons) = [2]
      pi(ifmin#) = [2]
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []

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: 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:

  (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:
  
      max/plus interpretations on natural numbers:
      
        le#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        s_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      le# = s
    
    partial status:
  
      pi(le#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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