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(x,nil())) -> x
  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))
  empty(nil()) -> true()
  empty(cons(n,x)) -> false()
  head(cons(n,x)) -> n
  tail(nil()) -> nil()
  tail(cons(n,x)) -> x
  |sort|(x) -> sortIter(x,nil())
  sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
  if(true(),x,y,z) -> y
  if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)


-- 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|#(x) -> sortIter#(x,nil())
p11: sortIter#(x,y) -> if#(empty(x),x,y,append(y,cons(min(x),nil())))
p12: sortIter#(x,y) -> empty#(x)
p13: sortIter#(x,y) -> min#(x)
p14: if#(false(),x,y,z) -> sortIter#(replace(min(x),head(x),tail(x)),z)
p15: if#(false(),x,y,z) -> replace#(min(x),head(x),tail(x))
p16: if#(false(),x,y,z) -> min#(x)
p17: if#(false(),x,y,z) -> head#(x)
p18: if#(false(),x,y,z) -> tail#(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(x,nil())) -> x
r9: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r10: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r11: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r12: replace(n,m,nil()) -> nil()
r13: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r14: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r15: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r16: empty(nil()) -> true()
r17: empty(cons(n,x)) -> false()
r18: head(cons(n,x)) -> n
r19: tail(nil()) -> nil()
r20: tail(cons(n,x)) -> x
r21: |sort|(x) -> sortIter(x,nil())
r22: sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
r23: if(true(),x,y,z) -> y
r24: if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)

The estimated dependency graph contains the following SCCs:

  {p11, p14}
  {p7, p9}
  {p1}
  {p3, p5, p6}
  {p2}


-- Reduction pair.

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

p1: sortIter#(x,y) -> if#(empty(x),x,y,append(y,cons(min(x),nil())))
p2: if#(false(),x,y,z) -> sortIter#(replace(min(x),head(x),tail(x)),z)

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(x,nil())) -> x
r9: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r10: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r11: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r12: replace(n,m,nil()) -> nil()
r13: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r14: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r15: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r16: empty(nil()) -> true()
r17: empty(cons(n,x)) -> false()
r18: head(cons(n,x)) -> n
r19: tail(nil()) -> nil()
r20: tail(cons(n,x)) -> x
r21: |sort|(x) -> sortIter(x,nil())
r22: sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
r23: if(true(),x,y,z) -> y
r24: if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)

The set of usable rules consists of

  r1, r2, r3, r4, r12, r13, r14, r15, r16, r17, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sortIter#_A(x1,x2) = 1
            if#_A(x1,x2,x3,x4) = 1
            empty_A(x1) = x1 + 43
            append_A(x1,x2) = max{114, x1 - 2}
            cons_A(x1,x2) = x2 + 17
            min_A(x1) = 62
            nil_A = 26
            false_A = 41
            replace_A(x1,x2,x3) = x3 + 2
            head_A(x1) = max{57, x1 - 36}
            tail_A(x1) = max{22, x1 + 20}
            eq_A(x1,x2) = 69
            |0|_A = 2
            true_A = 39
            s_A(x1) = max{2, x1 - 37}
            if_replace_A(x1,x2,x3,x4) = x4 + 2
            sortIter#!_A(x1,x2) = max{0, x2 - 116}
            if#!_A(x1,x2,x3,x4) = max{0, x3 - 116, x4 - 115}
            empty!_A(x1) = 30
            append!_A(x1,x2) = max{41, x1 - 73}
            cons!_A(x1,x2) = max{28, x2 + 13}
            min!_A(x1) = max{0, x1 - 22}
            nil!_A = 26
            false!_A = 29
            replace!_A(x1,x2,x3) = max{30, x3 + 1}
            head!_A(x1) = 0
            tail!_A(x1) = max{25, x1 - 26}
            eq!_A(x1,x2) = 30
            |0|!_A = 28
            true!_A = 27
            s!_A(x1) = x1 + 37
            if_replace!_A(x1,x2,x3,x4) = max{30, x1 - 40, x4 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sortIter#_A(x1,x2) = max{206, x2}
            if#_A(x1,x2,x3,x4) = max{206, x3 - 1, x4 + 205}
            empty_A(x1) = max{4, x1 - 36}
            append_A(x1,x2) = max{0, x2 - 139}
            cons_A(x1,x2) = x2 + 67
            min_A(x1) = 0
            nil_A = 1
            false_A = 29
            replace_A(x1,x2,x3) = x3 + 12
            head_A(x1) = x1 + 59
            tail_A(x1) = max{1, x1 - 57}
            eq_A(x1,x2) = 29
            |0|_A = 4
            true_A = 4
            s_A(x1) = x1 + 7
            if_replace_A(x1,x2,x3,x4) = x4 + 12
            sortIter#!_A(x1,x2) = x1 + 46
            if#!_A(x1,x2,x3,x4) = max{x1 + 41, x2 + 45, x4 - 2}
            empty!_A(x1) = max{0, x1 - 7}
            append!_A(x1,x2) = max{62, x1 - 1}
            cons!_A(x1,x2) = max{39, x2 + 1}
            min!_A(x1) = max{69, x1 + 47}
            nil!_A = 59
            false!_A = 60
            replace!_A(x1,x2,x3) = max{63, x3 - 2}
            head!_A(x1) = 60
            tail!_A(x1) = max{44, x1 - 2}
            eq!_A(x1,x2) = 65
            |0|!_A = 61
            true!_A = 58
            s!_A(x1) = max{66, x1 + 1}
            if_replace!_A(x1,x2,x3,x4) = max{39, x1 + 35, x4 - 30}
        
  
    partial status:
  
      pi(sortIter#) = []
      pi(if#) = []
      pi(empty) = [1]
      pi(append) = [1]
      pi(cons) = [2]
      pi(min) = [1]
      pi(nil) = []
      pi(false) = []
      pi(replace) = [3]
      pi(head) = []
      pi(tail) = [1]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(if_replace) = [4]

The following rules are removed:

  p1, p2

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(x,nil())) -> x
r9: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r10: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r11: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r12: replace(n,m,nil()) -> nil()
r13: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r14: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r15: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r16: empty(nil()) -> true()
r17: empty(cons(n,x)) -> false()
r18: head(cons(n,x)) -> n
r19: tail(nil()) -> nil()
r20: tail(cons(n,x)) -> x
r21: |sort|(x) -> sortIter(x,nil())
r22: sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
r23: if(true(),x,y,z) -> y
r24: if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r12, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_replace#_A(x1,x2,x3,x4) = max{0, x1 - 24, x2 - 3, x3 - 3, x4 - 10}
            false_A = 1
            cons_A(x1,x2) = max{8, x1 - 30, x2 + 2}
            replace#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 3, x3 - 10}
            eq_A(x1,x2) = max{23, x1 - 16, x2 - 16}
            |0|_A = 0
            true_A = 1
            s_A(x1) = x1 + 15
            le_A(x1,x2) = max{5, x1 - 16, x2 - 16}
            replace_A(x1,x2,x3) = max{12, x1 - 51, x2 - 5, x3 - 11}
            nil_A = 6
            tail_A(x1) = max{33, x1 - 7}
            if_replace#!_A(x1,x2,x3,x4) = max{35, x4 + 31}
            false!_A = 3
            cons!_A(x1,x2) = max{41, x1 - 68, x2 - 81}
            replace#!_A(x1,x2,x3) = max{36, x3 + 32}
            eq!_A(x1,x2) = max{42, x1 - 89, x2 - 114}
            |0|!_A = 4
            true!_A = 1
            s!_A(x1) = max{4, x1 - 113}
            le!_A(x1,x2) = max{6, x1 - 69, x2 - 112}
            replace!_A(x1,x2,x3) = max{42, x1 - 88, x2 - 91, x3 - 58}
            nil!_A = 5
            tail!_A(x1) = max{4, x1 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_replace#_A(x1,x2,x3,x4) = max{0, x1 - 19, x2 - 3, x3 - 3, x4 - 19}
            false_A = 1
            cons_A(x1,x2) = max{17, x1 - 9, x2 + 3}
            replace#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 3, x3 - 19}
            eq_A(x1,x2) = max{18, x1 - 9, x2 - 9}
            |0|_A = 0
            true_A = 1
            s_A(x1) = max{8, x1}
            le_A(x1,x2) = max{6, x1 - 9, x2 - 1}
            replace_A(x1,x2,x3) = max{21, x1 - 29, x2 - 11, x3 - 20}
            nil_A = 0
            tail_A(x1) = max{41, x1 - 4}
            if_replace#!_A(x1,x2,x3,x4) = max{4, x1 - 171, x2 - 2, x3 - 127, x4 - 144}
            false!_A = 4
            cons!_A(x1,x2) = max{8, x1 - 152, x2 - 134}
            replace#!_A(x1,x2,x3) = max{5, x1 - 1, x2 - 126, x3 - 142}
            eq!_A(x1,x2) = max{6, x1 - 130, x2 - 131}
            |0|!_A = 3
            true!_A = 2
            s!_A(x1) = max{1, x1 - 131}
            le!_A(x1,x2) = max{2, x1 - 153, x2 - 144}
            replace!_A(x1,x2,x3) = max{7, x1 - 253, x2 - 111, x3 - 123}
            nil!_A = 10
            tail!_A(x1) = max{0, x1 - 1}
        
  
    partial status:
  
      pi(if_replace#) = [2]
      pi(false) = []
      pi(cons) = [1, 2]
      pi(replace#) = [1]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(le) = [2]
      pi(replace) = []
      pi(nil) = []
      pi(tail) = [1]

The following rules are removed:

  p1, p2

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(x,nil())) -> x
r9: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r10: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r11: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r12: replace(n,m,nil()) -> nil()
r13: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r14: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r15: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r16: empty(nil()) -> true()
r17: empty(cons(n,x)) -> false()
r18: head(cons(n,x)) -> n
r19: tail(nil()) -> nil()
r20: tail(cons(n,x)) -> x
r21: |sort|(x) -> sortIter(x,nil())
r22: sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
r23: if(true(),x,y,z) -> y
r24: if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)

The set of usable rules consists of

  r5, r6, r12, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
            s_A(x1) = max{2, x1}
            le_A(x1,x2) = max{3, x1 - 3, x2 - 7}
            |0|_A = 2
            true_A = 1
            false_A = 1
            replace_A(x1,x2,x3) = max{16, x1 - 7, x2 - 12, x3 - 5}
            nil_A = 0
            tail_A(x1) = max{21, x1 - 1}
            eq#!_A(x1,x2) = max{9, x1 - 1, x2 - 3}
            s!_A(x1) = 8
            le!_A(x1,x2) = max{9, x1 - 37, x2 - 54}
            |0|!_A = 6
            true!_A = 1
            false!_A = 5
            replace!_A(x1,x2,x3) = max{4, x1 - 33, x2 - 38, x3 - 23}
            nil!_A = 3
            tail!_A(x1) = max{4, x1 - 4}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = max{0, x1 - 5, x2 - 5}
            s_A(x1) = x1 + 4
            le_A(x1,x2) = max{3, x1 - 5, x2 - 7}
            |0|_A = 4
            true_A = 1
            false_A = 1
            replace_A(x1,x2,x3) = max{13, x1 - 12, x2 - 12, x3 - 13}
            nil_A = 0
            tail_A(x1) = max{24, x1 - 11}
            eq#!_A(x1,x2) = max{2, x1 - 1, x2 - 77}
            s!_A(x1) = max{1, x1 - 77}
            le!_A(x1,x2) = max{4, x1 - 76, x2 - 90}
            |0|!_A = 7
            true!_A = 3
            false!_A = 6
            replace!_A(x1,x2,x3) = max{5, x1 - 50, x2 - 50, x3 - 44}
            nil!_A = 4
            tail!_A(x1) = max{5, x1 - 55}
        
  
    partial status:
  
      pi(eq#) = [2]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(replace) = []
      pi(nil) = []
      pi(tail) = []

The following rules are removed:

  p1

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(x,nil())) -> x
r9: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r10: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r11: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r12: replace(n,m,nil()) -> nil()
r13: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r14: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r15: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r16: empty(nil()) -> true()
r17: empty(cons(n,x)) -> false()
r18: head(cons(n,x)) -> n
r19: tail(nil()) -> nil()
r20: tail(cons(n,x)) -> x
r21: |sort|(x) -> sortIter(x,nil())
r22: sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
r23: if(true(),x,y,z) -> y
r24: if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)

The set of usable rules consists of

  r5, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_min#_A(x1,x2) = max{0, x1 - 107, x2 - 73}
            false_A = 1
            cons_A(x1,x2) = max{58, x1 + 16, x2 + 13}
            min#_A(x1) = max{0, x1 - 73}
            le_A(x1,x2) = max{106, x1 - 2, x2 - 2}
            true_A = 91
            |0|_A = 0
            s_A(x1) = x1 + 1
            if_min#!_A(x1,x2) = max{40, x1 - 92, x2 - 44}
            false!_A = 6
            cons!_A(x1,x2) = max{x1 + 12, x2 - 3}
            min#!_A(x1) = max{11, x1 - 30}
            le!_A(x1,x2) = max{8, x1 - 15, x2 - 2}
            true!_A = 2
            |0|!_A = 3
            s!_A(x1) = max{7, x1 - 103}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_min#_A(x1,x2) = max{0, x1 - 31, x2 - 14}
            false_A = 7
            cons_A(x1,x2) = max{x1 + 8, x2 + 4}
            min#_A(x1) = max{0, x1 - 14}
            le_A(x1,x2) = max{30, x1 - 13, x2 - 13}
            true_A = 1
            |0|_A = 1
            s_A(x1) = max{12, x1 + 1}
            if_min#!_A(x1,x2) = max{0, x1 - 30}
            false!_A = 4
            cons!_A(x1,x2) = 8
            min#!_A(x1) = max{0, x1 - 26}
            le!_A(x1,x2) = max{9, x1 - 2, x2 - 13}
            true!_A = 2
            |0|!_A = 3
            s!_A(x1) = max{4, x1 - 14}
        
  
    partial status:
  
      pi(if_min#) = [1, 2]
      pi(false) = []
      pi(cons) = [1, 2]
      pi(min#) = [1]
      pi(le) = [1]
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]

The following rules are removed:

  p1, p2, p3

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(x,nil())) -> x
r9: min(cons(n,cons(m,x))) -> if_min(le(n,m),cons(n,cons(m,x)))
r10: if_min(true(),cons(n,cons(m,x))) -> min(cons(n,x))
r11: if_min(false(),cons(n,cons(m,x))) -> min(cons(m,x))
r12: replace(n,m,nil()) -> nil()
r13: replace(n,m,cons(k,x)) -> if_replace(eq(n,k),n,m,cons(k,x))
r14: if_replace(true(),n,m,cons(k,x)) -> cons(m,x)
r15: if_replace(false(),n,m,cons(k,x)) -> cons(k,replace(n,m,x))
r16: empty(nil()) -> true()
r17: empty(cons(n,x)) -> false()
r18: head(cons(n,x)) -> n
r19: tail(nil()) -> nil()
r20: tail(cons(n,x)) -> x
r21: |sort|(x) -> sortIter(x,nil())
r22: sortIter(x,y) -> if(empty(x),x,y,append(y,cons(min(x),nil())))
r23: if(true(),x,y,z) -> y
r24: if(false(),x,y,z) -> sortIter(replace(min(x),head(x),tail(x)),z)

The set of usable rules consists of

  r5, r6, r12, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
            s_A(x1) = max{2, x1}
            le_A(x1,x2) = max{5, x1 - 3, x2 - 25}
            |0|_A = 0
            true_A = 1
            false_A = 3
            replace_A(x1,x2,x3) = max{13, x1 - 22, x2 - 32, x3 - 27}
            nil_A = 4
            tail_A(x1) = max{51, x1 - 5}
            le#!_A(x1,x2) = max{4, x1 - 1, x2 - 3}
            s!_A(x1) = max{3, x1 - 4}
            le!_A(x1,x2) = max{7, x1 - 79, x2 - 93}
            |0|!_A = 7
            true!_A = 6
            false!_A = 2
            replace!_A(x1,x2,x3) = max{6, x1 - 66, x2 - 96, x3 - 89}
            nil!_A = 0
            tail!_A(x1) = max{1, x1 - 69}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
            s_A(x1) = max{11, x1 + 1}
            le_A(x1,x2) = max{3, x1 - 22, x2 - 18}
            |0|_A = 17
            true_A = 1
            false_A = 1
            replace_A(x1,x2,x3) = max{22, x1 - 22, x2 - 6, x3 - 5}
            nil_A = 0
            tail_A(x1) = max{26, x1 - 3}
            le#!_A(x1,x2) = max{9, x1 - 12, x2 - 2}
            s!_A(x1) = max{5, x1 - 3}
            le!_A(x1,x2) = max{6, x1 - 4, x2 - 18}
            |0|!_A = 5
            true!_A = 2
            false!_A = 4
            replace!_A(x1,x2,x3) = max{5, x1 - 71, x2 - 56, x3 - 27}
            nil!_A = 0
            tail!_A(x1) = max{1, x1 - 30}
        
  
    partial status:
  
      pi(le#) = [2]
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(replace) = []
      pi(nil) = []
      pi(tail) = []

The following rules are removed:

  p1

Then no dependency pair remains.