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, r7, r12, r13, r14, r15, r16, r17, r19, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          sortIter#_A(x1,x2) = 4
          if#_A(x1,x2,x3,x4) = 4
          empty_A(x1) = 7
          append_A(x1,x2) = 1
          cons_A(x1,x2) = x2
          min_A(x1) = x1
          nil_A() = 1
          false_A() = 5
          replace_A(x1,x2,x3) = x3 + 5
          head_A(x1) = 3
          tail_A(x1) = x1 + 1
          eq_A(x1,x2) = 5
          |0|_A() = 6
          true_A() = 0
          s_A(x1) = 6
          le_A(x1,x2) = 5
          if_replace_A(x1,x2,x3,x4) = x1 + x4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sortIter#_A(x1,x2) = 52
          if#_A(x1,x2,x3,x4) = max{52, x4 + 51}
          empty_A(x1) = max{1, x1 - 2}
          append_A(x1,x2) = max{0, x2 - 30}
          cons_A(x1,x2) = max{22, x2 + 8}
          min_A(x1) = 7
          nil_A = 0
          false_A = 10
          replace_A(x1,x2,x3) = x3 + 2
          head_A(x1) = max{2, x1 - 2}
          tail_A(x1) = max{0, x1 - 6}
          eq_A(x1,x2) = 35
          |0|_A = 3
          true_A = 1
          s_A(x1) = max{3, x1 + 1}
          le_A(x1,x2) = max{19, x1 - 4, x2 - 4}
          if_replace_A(x1,x2,x3,x4) = max{21, x1 - 11, x4 + 2}
          sortIter#!_A(x1,x2) = max{8, x1}
          if#!_A(x1,x2,x3,x4) = max{x1 - 2, x2 - 4, x4 + 7}
          empty!_A(x1) = 55
          append!_A(x1,x2) = 28
          cons!_A(x1,x2) = max{52, x2 - 17}
          min!_A(x1) = max{53, x1 - 30}
          nil!_A = 55
          false!_A = 56
          replace!_A(x1,x2,x3) = max{57, x3 + 6}
          head!_A(x1) = max{52, x1 - 4}
          tail!_A(x1) = max{51, x1 + 1}
          eq!_A(x1,x2) = max{x1 + 58, x2 + 58}
          |0|!_A = 57
          true!_A = 54
          s!_A(x1) = x1 + 57
          le!_A(x1,x2) = 58
          if_replace!_A(x1,x2,x3,x4) = max{52, x4 - 8}
  
    partial status:
  
      pi(sortIter#) = []
      pi(if#) = []
      pi(empty) = []
      pi(append) = []
      pi(cons) = []
      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



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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

  r2, r3, r5, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_replace#_A(x1,x2,x3,x4) = 1
          false_A() = 6
          cons_A(x1,x2) = x1 + 7
          replace#_A(x1,x2,x3) = 1
          eq_A(x1,x2) = x1 + x2 + 2
          |0|_A() = 7
          s_A(x1) = x1 + 7
          le_A(x1,x2) = 2
          true_A() = 1
          tail_A(x1) = 7
          nil_A() = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_replace#_A(x1,x2,x3,x4) = max{1, x2 - 21, x3 - 21, x4 - 28}
          false_A = 2
          cons_A(x1,x2) = max{x1 - 13, x2 + 27}
          replace#_A(x1,x2,x3) = max{1, x1 - 21, x2 - 21, x3 - 28}
          eq_A(x1,x2) = max{16, x1 - 11}
          |0|_A = 1
          s_A(x1) = max{0, x1 - 18}
          le_A(x1,x2) = max{65, x1 - 2, x2 - 2}
          true_A = 17
          tail_A(x1) = max{60, x1 - 70}
          nil_A = 0
          if_replace#!_A(x1,x2,x3,x4) = max{17, x4 + 11}
          false!_A = 44
          cons!_A(x1,x2) = 46
          replace#!_A(x1,x2,x3) = max{18, x3 + 12}
          eq!_A(x1,x2) = x2 + 44
          |0|!_A = 45
          s!_A(x1) = 44
          le!_A(x1,x2) = 44
          true!_A = 42
          tail!_A(x1) = max{73, x1 - 1}
          nil!_A = 70
  
    partial status:
  
      pi(if_replace#) = []
      pi(false) = []
      pi(cons) = [1]
      pi(replace#) = []
      pi(eq) = [1]
      pi(|0|) = []
      pi(s) = [1]
      pi(le) = []
      pi(true) = []
      pi(tail) = []
      pi(nil) = []

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

  r1, r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          eq#_A(x1,x2) = x1 + x2 + 8
          s_A(x1) = x1 + 7
          eq_A(x1,x2) = 6
          |0|_A() = 6
          true_A() = 2
          false_A() = 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          eq#_A(x1,x2) = max{0, x1 - 111, x2 - 111}
          s_A(x1) = max{120, x1 + 110}
          eq_A(x1,x2) = max{34, x1 - 121, x2 - 121}
          |0|_A = 120
          true_A = 1
          false_A = 1
          eq#!_A(x1,x2) = 32
          s!_A(x1) = x1 + 31
          eq!_A(x1,x2) = max{22, x2 - 87}
          |0|!_A = 33
          true!_A = 18
          false!_A = 32
  
    partial status:
  
      pi(eq#) = [2]
      pi(s) = [1]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []

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

  r1, r2, r3, r5, r6, r7, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_min#_A(x1,x2) = x2 + 1
          false_A() = 2
          cons_A(x1,x2) = 10
          min#_A(x1) = x1 + 1
          le_A(x1,x2) = 9
          true_A() = 2
          eq_A(x1,x2) = 7
          |0|_A() = 3
          s_A(x1) = x1 + 8
          replace_A(x1,x2,x3) = 6
          nil_A() = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_min#_A(x1,x2) = 1
          false_A = 21
          cons_A(x1,x2) = max{x1 + 30, x2 + 30}
          min#_A(x1) = 1
          le_A(x1,x2) = max{19, x1 - 18, x2 + 1}
          true_A = 19
          eq_A(x1,x2) = max{147, x1 - 40, x2 - 40}
          |0|_A = 19
          s_A(x1) = max{39, x1 + 17}
          replace_A(x1,x2,x3) = max{125, x1 + 63, x2 + 63, x3 + 94}
          nil_A = 64
          if_min#!_A(x1,x2) = max{6, x1 - 12, x2 - 60}
          false!_A = 6
          cons!_A(x1,x2) = 5
          min#!_A(x1) = max{7, x1 - 60}
          le!_A(x1,x2) = 2
          true!_A = 1
          eq!_A(x1,x2) = 4
          |0|!_A = 7
          s!_A(x1) = max{7, x1 + 1}
          replace!_A(x1,x2,x3) = max{3, x1 - 219, x2 - 257, x3 - 226}
          nil!_A = 2
  
    partial status:
  
      pi(if_min#) = [2]
      pi(false) = []
      pi(cons) = []
      pi(min#) = [1]
      pi(le) = []
      pi(true) = []
      pi(eq) = []
      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

  r1, r2, r3, r5, r6, r7, r12, r17

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          min#_A(x1) = x1 + 4
          cons_A(x1,x2) = x1 + x2 + 4
          if_min#_A(x1,x2) = x1 + x2
          le_A(x1,x2) = 4
          true_A() = 0
          eq_A(x1,x2) = 5
          |0|_A() = 1
          s_A(x1) = x1 + 6
          false_A() = 3
          replace_A(x1,x2,x3) = 6
          nil_A() = 0
          empty_A(x1) = x1 + 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          min#_A(x1) = max{8, x1 - 55}
          cons_A(x1,x2) = max{62, x1 + 33}
          if_min#_A(x1,x2) = max{8, x2 - 55}
          le_A(x1,x2) = x2 + 1
          true_A = 1
          eq_A(x1,x2) = max{x1 + 94, x2 - 163}
          |0|_A = 118
          s_A(x1) = max{162, x1 + 118}
          false_A = 2
          replace_A(x1,x2,x3) = max{67, x1 + 24, x2 - 22, x3 + 26}
          nil_A = 62
          empty_A(x1) = max{8, x1 - 63}
          min#!_A(x1) = max{1, x1 - 44}
          cons!_A(x1,x2) = max{6, x1 - 1, x2 - 25}
          if_min#!_A(x1,x2) = max{2, x2 - 55}
          le!_A(x1,x2) = max{3, x1 - 376}
          true!_A = 41
          eq!_A(x1,x2) = max{42, x1 - 119, x2 - 376}
          |0|!_A = 4
          s!_A(x1) = max{4, x1 - 375}
          false!_A = 5
          replace!_A(x1,x2,x3) = max{43, x1 - 118, x2 - 523, x3 - 408}
          nil!_A = 0
          empty!_A(x1) = max{3, x1 - 22}
  
    partial status:
  
      pi(min#) = []
      pi(cons) = []
      pi(if_min#) = [2]
      pi(le) = []
      pi(true) = []
      pi(eq) = []
      pi(|0|) = []
      pi(s) = [1]
      pi(false) = []
      pi(replace) = []
      pi(nil) = []
      pi(empty) = [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: 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          le#_A(x1,x2) = x1 + x2 + 10
          s_A(x1) = x1 + 9
  
    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{21, x1 + 1}
          le#!_A(x1,x2) = max{22, x1 - 24, x2 - 22}
          s!_A(x1) = max{21, x1 - 23}
  
    partial status:
  
      pi(le#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.