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:
  
      max/plus interpretation on natural numbers:
      
        sortIter#_A(x1,x2) = x1 + 18
        if#_A(x1,x2,x3,x4) = max{x1 - 4, x2 + 17}
        empty_A(x1) = x1 + 5
        append_A(x1,x2) = max{59, x1 + 58, x2 + 4}
        cons_A(x1,x2) = max{53, x2 + 39}
        min_A(x1) = 2
        nil_A = 15
        false_A = 45
        replace_A(x1,x2,x3) = x3 + 1
        head_A(x1) = 18
        tail_A(x1) = max{15, x1 - 38}
        eq_A(x1,x2) = 48
        |0|_A = 49
        true_A = 4
        s_A(x1) = x1 + 49
        if_replace_A(x1,x2,x3,x4) = max{54, x1 - 46, x4 + 1}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            sortIter#_A(x1,x2) = 0
            if#_A(x1,x2,x3,x4) = 0
            empty_A(x1) = x1 + 1
            append_A(x1,x2) = x1 + 1
            cons_A(x1,x2) = x2 + 2
            min_A(x1) = 1
            nil_A() = 1
            false_A() = 3
            replace_A(x1,x2,x3) = x3 + 3
            head_A(x1) = x1 + 1
            tail_A(x1) = x1
            eq_A(x1,x2) = 3
            |0|_A() = 0
            true_A() = 2
            s_A(x1) = x1 + 1
            if_replace_A(x1,x2,x3,x4) = x1 + x4
            sortIter#!_A(x1,x2) = x2 + 2
            if#!_A(x1,x2,x3,x4) = x1
            empty!_A(x1) = 1
            append!_A(x1,x2) = 0
            cons!_A(x1,x2) = 6
            min!_A(x1) = 4
            nil!_A() = 6
            false!_A() = 10
            replace!_A(x1,x2,x3) = 8
            head!_A(x1) = 2
            tail!_A(x1) = 9
            eq!_A(x1,x2) = x1 + x2 + 4
            |0|!_A() = 11
            true!_A() = 5
            s!_A(x1) = x1 + 10
            if_replace!_A(x1,x2,x3,x4) = 7
  
    partial status:
  
      pi(sortIter#) = []
      pi(if#) = []
      pi(empty) = []
      pi(append) = []
      pi(cons) = []
      pi(min) = []
      pi(nil) = []
      pi(false) = []
      pi(replace) = []
      pi(head) = []
      pi(tail) = []
      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

  r3

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{2, x4 - 14}
        false_A = 14
        cons_A(x1,x2) = max{14, x1 + 5, x2 + 4}
        replace#_A(x1,x2,x3) = max{1, x3 - 11}
        eq_A(x1,x2) = max{14, x1 - 2}
        s_A(x1) = x1 + 52
        |0|_A = 15
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            if_replace#_A(x1,x2,x3,x4) = 0
            false_A() = 5
            cons_A(x1,x2) = x1 + x2
            replace#_A(x1,x2,x3) = 0
            eq_A(x1,x2) = x2 + 3
            s_A(x1) = x1 + 1
            |0|_A() = 2
            if_replace#!_A(x1,x2,x3,x4) = x2 + x3 + x4 + 5
            false!_A() = 3
            cons!_A(x1,x2) = x2 + 4
            replace#!_A(x1,x2,x3) = x3 + 1
            eq!_A(x1,x2) = x2
            s!_A(x1) = 1
            |0|!_A() = 4
  
    partial status:
  
      pi(if_replace#) = []
      pi(false) = []
      pi(cons) = [1]
      pi(replace#) = []
      pi(eq) = []
      pi(s) = [1]
      pi(|0|) = []

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

  r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        eq#_A(x1,x2) = max{31, x2 - 1}
        s_A(x1) = max{35, x1 + 33}
        min_A(x1) = max{30, x1 + 24}
        cons_A(x1,x2) = max{x1 + 36, x2 - 7}
        nil_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            eq#_A(x1,x2) = x1 + x2
            s_A(x1) = x1 + 1
            min_A(x1) = x1 + 1
            cons_A(x1,x2) = x1 + 3
            nil_A() = 2
            eq#!_A(x1,x2) = x2
            s!_A(x1) = x1 + 9
            min!_A(x1) = 7
            cons!_A(x1,x2) = x1 + 9
            nil!_A() = 2
  
    partial status:
  
      pi(eq#) = []
      pi(s) = [1]
      pi(min) = [1]
      pi(cons) = [1]
      pi(nil) = []

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

  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) = max{0, x2 - 7}
        false_A = 0
        cons_A(x1,x2) = 6
        min#_A(x1) = 0
        le_A(x1,x2) = 0
        true_A = 0
        replace_A(x1,x2,x3) = max{8, x1 - 4, x2 - 9}
        nil_A = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            if_min#_A(x1,x2) = x2
            false_A() = 4
            cons_A(x1,x2) = x1 + x2 + 4
            min#_A(x1) = x1
            le_A(x1,x2) = 1
            true_A() = 4
            replace_A(x1,x2,x3) = x1 + x2 + x3 + 4
            nil_A() = 2
            if_min#!_A(x1,x2) = x2
            false!_A() = 7
            cons!_A(x1,x2) = x1 + x2 + 5
            min#!_A(x1) = x1 + 2
            le!_A(x1,x2) = x1 + 9
            true!_A() = 7
            replace!_A(x1,x2,x3) = x2 + 9
            nil!_A() = 8
  
    partial status:
  
      pi(if_min#) = []
      pi(false) = []
      pi(cons) = []
      pi(min#) = []
      pi(le) = []
      pi(true) = []
      pi(replace) = []
      pi(nil) = []

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = max{25, x2 - 1}
        s_A(x1) = max{27, x1 + 24}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            le#_A(x1,x2) = x1 + x2
            s_A(x1) = x1 + 1
            le#!_A(x1,x2) = 0
            s!_A(x1) = x1 + 12
  
    partial status:
  
      pi(le#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.