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, r5, r6, r12, r13, r14, r15, r16, r17, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sortIter#_A(x1,x2) = max{18, x1 - 33, x2 - 1}
        if#_A(x1,x2,x3,x4) = max{x1 - 43, x2 - 34, x4 + 1}
        empty_A(x1) = max{60, x1 + 9}
        append_A(x1,x2) = max{0, x2 - 52}
        cons_A(x1,x2) = x2 + 54
        min_A(x1) = max{107, x1 - 55}
        nil_A = 14
        false_A = 62
        replace_A(x1,x2,x3) = x3 + 12
        head_A(x1) = 61
        tail_A(x1) = max{20, x1 - 14}
        eq_A(x1,x2) = 68
        |0|_A = 63
        true_A = 59
        s_A(x1) = max{69, x1 - 57}
        le_A(x1,x2) = max{167, x1 - 64, x2 - 70}
        if_replace_A(x1,x2,x3,x4) = max{51, x1 - 58, x4 + 12}
  
    base order 2:
  
      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) = 7
          append_A(x1,x2) = max{x1 - 18, x2 + 16}
          cons_A(x1,x2) = max{12, x2}
          min_A(x1) = x1 + 10
          nil_A = 14
          false_A = 5
          replace_A(x1,x2,x3) = x3 + 11
          head_A(x1) = x1 + 9
          tail_A(x1) = max{1, x1}
          eq_A(x1,x2) = max{4, x1 - 1, x2 + 1}
          |0|_A = 2
          true_A = 4
          s_A(x1) = max{6, x1}
          le_A(x1,x2) = max{5, x1 - 7, x2 - 7}
          if_replace_A(x1,x2,x3,x4) = x4 + 11
          sortIter#!_A(x1,x2) = x2 + 20
          if#!_A(x1,x2,x3,x4) = max{16, x3 + 1}
          empty!_A(x1) = max{21, x1 + 10}
          append!_A(x1,x2) = max{x1 + 19, x2}
          cons!_A(x1,x2) = 0
          min!_A(x1) = 17
          nil!_A = 19
          false!_A = 22
          replace!_A(x1,x2,x3) = 0
          head!_A(x1) = max{15, x1 - 1}
          tail!_A(x1) = max{23, x1 - 15}
          eq!_A(x1,x2) = max{23, x1 - 7, x2 - 9}
          |0|!_A = 25
          true!_A = 24
          s!_A(x1) = max{0, x1 - 8}
          le!_A(x1,x2) = 0
          if_replace!_A(x1,x2,x3,x4) = 0
  
    partial status:
  
      pi(sortIter#) = []
      pi(if#) = []
      pi(empty) = [1]
      pi(append) = []
      pi(cons) = [2]
      pi(min) = []
      pi(nil) = []
      pi(false) = []
      pi(replace) = [3]
      pi(head) = []
      pi(tail) = []
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = []
      pi(le) = []
      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, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_replace#_A(x1,x2,x3,x4) = max{9, x3 + 1}
        false_A = 1
        cons_A(x1,x2) = max{8, x1 - 32, x2 - 32}
        replace#_A(x1,x2,x3) = max{9, x2 + 1}
        eq_A(x1,x2) = max{35, x1 - 33, x2 - 33}
        |0|_A = 20
        true_A = 9
        s_A(x1) = max{20, x1}
        le_A(x1,x2) = max{19, x1 - 21, x2 - 21}
        empty_A(x1) = max{31, x1 - 31}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_replace#_A(x1,x2,x3,x4) = max{0, x1 - 21, x2 - 31, x3 - 8, x4 - 8}
          false_A = 1
          cons_A(x1,x2) = max{x1 + 4, x2 + 6}
          replace#_A(x1,x2,x3) = max{0, x1 - 31, x2 - 8, x3 - 8}
          eq_A(x1,x2) = max{20, x1 - 10, x2 - 10}
          |0|_A = 0
          true_A = 0
          s_A(x1) = max{9, x1 + 1}
          le_A(x1,x2) = max{14, x1 - 10, x2 - 2}
          empty_A(x1) = max{119, x1 - 14}
          if_replace#!_A(x1,x2,x3,x4) = max{1, x4 - 3}
          false!_A = 1
          cons!_A(x1,x2) = max{5, x1 - 115, x2 - 185}
          replace#!_A(x1,x2,x3) = max{2, x3 - 2}
          eq!_A(x1,x2) = max{6, x1 - 10, x2 - 2}
          |0|!_A = 2
          true!_A = 1
          s!_A(x1) = max{1, x1 - 1}
          le!_A(x1,x2) = max{6, x1 - 116, x2 - 106}
          empty!_A(x1) = max{2, x1 - 131}
  
    partial status:
  
      pi(if_replace#) = [3]
      pi(false) = []
      pi(cons) = []
      pi(replace#) = [2]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [1]
      pi(le) = []
      pi(empty) = []

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

  r6, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = 1
        s_A(x1) = max{6, x1 - 73}
        le_A(x1,x2) = max{104, x1 - 7, x2 - 72}
        |0|_A = 6
        false_A = 1
        empty_A(x1) = max{33, x1 - 32}
        cons_A(x1,x2) = max{4, x1 - 14, x2 - 57}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          eq#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
          s_A(x1) = max{6, x1 + 1}
          le_A(x1,x2) = max{67, x1 - 7, x2 - 2}
          |0|_A = 0
          false_A = 3
          empty_A(x1) = max{67, x1 - 122}
          cons_A(x1,x2) = max{46, x2 - 72}
          eq#!_A(x1,x2) = max{3, x1 - 2, x2 - 7}
          s!_A(x1) = max{2, x1 - 8}
          le!_A(x1,x2) = max{6, x1 - 7, x2 - 9}
          |0|!_A = 2
          false!_A = 1
          empty!_A(x1) = max{7, x1 - 122}
          cons!_A(x1,x2) = max{5, x1 - 120, x2 - 122}
  
    partial status:
  
      pi(eq#) = []
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(false) = []
      pi(empty) = []
      pi(cons) = []

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, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_min#_A(x1,x2) = 41
        false_A = 69
        cons_A(x1,x2) = max{40, x1, x2 - 24}
        min#_A(x1) = 41
        le_A(x1,x2) = max{x1 + 48, x2 - 23}
        true_A = 1
        |0|_A = 20
        s_A(x1) = x1 + 70
        replace_A(x1,x2,x3) = max{23, x1 - 26, x2 - 84}
        nil_A = 17
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_min#_A(x1,x2) = max{0, x1 - 18, x2 - 6}
          false_A = 10
          cons_A(x1,x2) = max{x1 - 23, x2 + 2}
          min#_A(x1) = max{0, x1 - 6}
          le_A(x1,x2) = max{17, x1 - 11, x2 - 11}
          true_A = 1
          |0|_A = 0
          s_A(x1) = max{61, x1 + 2}
          replace_A(x1,x2,x3) = max{35, x1 - 12, x2 - 41, x3 - 41}
          nil_A = 0
          if_min#!_A(x1,x2) = max{57, x2 + 54}
          false!_A = 31
          cons!_A(x1,x2) = max{29, x1 - 55, x2 + 25}
          min#!_A(x1) = x1 + 54
          le!_A(x1,x2) = max{1, x1 - 59, x2 - 1}
          true!_A = 27
          |0|!_A = 30
          s!_A(x1) = max{32, x1 - 63}
          replace!_A(x1,x2,x3) = max{32, x1 - 170, x2 - 195, x3 - 109}
          nil!_A = 28
  
    partial status:
  
      pi(if_min#) = []
      pi(false) = []
      pi(cons) = []
      pi(min#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(replace) = []
      pi(nil) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: min#(cons(n,cons(m,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))
p2: 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 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,x))) -> if_min#(le(n,m),cons(n,cons(m,x)))
p2: 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, r17, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        min#_A(x1) = max{2, x1 - 36}
        cons_A(x1,x2) = max{36, x2 + 18}
        if_min#_A(x1,x2) = max{1, x2 - 52}
        le_A(x1,x2) = x2 + 192
        true_A = 5
        |0|_A = 9
        s_A(x1) = x1 + 2
        false_A = 0
        empty_A(x1) = max{56, x1 - 167}
        tail_A(x1) = max{36, x1 - 17}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          min#_A(x1) = max{0, x1 - 88}
          cons_A(x1,x2) = max{x1 - 2, x2 + 17}
          if_min#_A(x1,x2) = max{0, x1 - 129, x2 - 88}
          le_A(x1,x2) = 57
          true_A = 57
          |0|_A = 4
          s_A(x1) = x1 + 2
          false_A = 55
          empty_A(x1) = max{60, x1 + 37}
          tail_A(x1) = max{15, x1 - 17}
          min#!_A(x1) = 7
          cons!_A(x1,x2) = max{7, x1 - 52, x2 - 71}
          if_min#!_A(x1,x2) = max{1, x1 - 50}
          le!_A(x1,x2) = max{3, x1 - 3, x2 + 1}
          true!_A = 8
          |0|!_A = 9
          s!_A(x1) = 11
          false!_A = 1
          empty!_A(x1) = max{12, x1 - 18}
          tail!_A(x1) = max{10, x1 - 17}
  
    partial status:
  
      pi(min#) = []
      pi(cons) = []
      pi(if_min#) = []
      pi(le) = []
      pi(true) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []
      pi(empty) = []
      pi(tail) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 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(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

  r2, r3, r6, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = 1
        s_A(x1) = max{2, x1 - 77}
        eq_A(x1,x2) = max{14, x1 - 5, x2 - 5}
        |0|_A = 4
        false_A = 1
        le_A(x1,x2) = max{14, x1 - 42, x2 - 76}
        empty_A(x1) = max{58, x1 - 14}
        cons_A(x1,x2) = max{13, x1 - 42, x2 - 48}
  
    base order 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{5, x1 + 1}
          eq_A(x1,x2) = max{41, x1 - 6, x2 - 6}
          |0|_A = 5
          false_A = 1
          le_A(x1,x2) = max{41, x1 - 6, x2 - 6}
          empty_A(x1) = max{53, x1 - 18}
          cons_A(x1,x2) = max{3, x2 - 30}
          le#!_A(x1,x2) = max{2, x1 - 2, x2 - 6}
          s!_A(x1) = max{1, x1 - 18}
          eq!_A(x1,x2) = max{70, x1 - 18, x2 - 6}
          |0|!_A = 2
          false!_A = 1
          le!_A(x1,x2) = max{70, x1 - 17, x2 - 6}
          empty!_A(x1) = max{2, x1 - 18}
          cons!_A(x1,x2) = max{69, x1 - 16, x2 - 18}
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(eq) = []
      pi(|0|) = []
      pi(false) = []
      pi(le) = []
      pi(empty) = []
      pi(cons) = []

The following rules are removed:

  p1

Then no dependency pair remains.