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, 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!):
        
          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) = 4
              append_A(x1,x2) = x1 + 1
              cons_A(x1,x2) = x2 + 1
              min_A(x1) = 1
              nil_A() = 1
              false_A() = 4
              replace_A(x1,x2,x3) = x3 + 1
              head_A(x1) = x1 + 1
              tail_A(x1) = x1 + 1
              eq_A(x1,x2) = 3
              |0|_A() = 1
              true_A() = 3
              if_replace_A(x1,x2,x3,x4) = x4 + 1
              sortIter#!_A(x1,x2) = 1
              if#!_A(x1,x2,x3,x4) = 1
              empty!_A(x1) = 0
              append!_A(x1,x2) = 9
              cons!_A(x1,x2) = 5
              min!_A(x1) = 2
              nil!_A() = 4
              false!_A() = 3
              replace!_A(x1,x2,x3) = x3 + 6
              head!_A(x1) = 3
              tail!_A(x1) = 8
              eq!_A(x1,x2) = 4
              |0|!_A() = 4
              true!_A() = 3
              if_replace!_A(x1,x2,x3,x4) = x4 + 5
        
        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) = max{0, x1 - 3}
            append_A(x1,x2) = 5
            cons_A(x1,x2) = max{91, x2 + 33}
            min_A(x1) = 3
            nil_A = 4
            false_A = 49
            replace_A(x1,x2,x3) = max{40, x3}
            head_A(x1) = max{4, x1 - 190}
            tail_A(x1) = max{4, x1 - 33}
            eq_A(x1,x2) = max{0, x1 - 90, x2 - 139}
            |0|_A = 91
            true_A = 1
            if_replace_A(x1,x2,x3,x4) = max{91, x4}
            sortIter#!_A(x1,x2) = max{x1 + 121, x2 + 114}
            if#!_A(x1,x2,x3,x4) = max{x1 + 113, x2 + 89, x4 + 115}
            empty!_A(x1) = 49
            append!_A(x1,x2) = max{122, x1 + 8}
            cons!_A(x1,x2) = max{x1 + 152, x2 + 92}
            min!_A(x1) = max{47, x1 - 37}
            nil!_A = 8
            false!_A = 50
            replace!_A(x1,x2,x3) = max{49, x3 - 2}
            head!_A(x1) = max{4, x1 - 37}
            tail!_A(x1) = 6
            eq!_A(x1,x2) = max{x1 - 90, x2 + 153}
            |0|!_A = 52
            true!_A = 53
            if_replace!_A(x1,x2,x3,x4) = max{x2 + 91, x3 + 153, x4 + 60}
        
  
    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) = [1]
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      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

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_replace#_A(x1,x2,x3,x4) = x4
              false_A() = 3
              cons_A(x1,x2) = x1 + x2 + 6
              replace#_A(x1,x2,x3) = x3
              eq_A(x1,x2) = x1 + 2
              |0|_A() = 1
              true_A() = 3
              s_A(x1) = x1 + 1
              if_replace#!_A(x1,x2,x3,x4) = x4 + 2
              false!_A() = 4
              cons!_A(x1,x2) = x1 + x2 + 5
              replace#!_A(x1,x2,x3) = x3 + 3
              eq!_A(x1,x2) = 8
              |0|!_A() = 7
              true!_A() = 6
              s!_A(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 - 61, x2 - 3, x3 - 3, x4 - 37}
            false_A = 1
            cons_A(x1,x2) = max{35, x1 + 1, x2}
            replace#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 3, x3 - 37}
            eq_A(x1,x2) = max{60, x1 - 19, x2 - 101}
            |0|_A = 4
            true_A = 13
            s_A(x1) = max{18, x1 + 1}
            if_replace#!_A(x1,x2,x3,x4) = max{46, x1 - 177, x2 - 197, x3 - 1, x4 - 233}
            false!_A = 13
            cons!_A(x1,x2) = max{45, x1 + 11, x2 - 116}
            replace#!_A(x1,x2,x3) = max{47, x1 - 197, x3 - 116}
            eq!_A(x1,x2) = max{15, x1 - 2, x2 - 116}
            |0|!_A = 14
            true!_A = 3
            s!_A(x1) = x1 + 14
        
  
    partial status:
  
      pi(if_replace#) = []
      pi(false) = []
      pi(cons) = [1]
      pi(replace#) = []
      pi(eq) = []
      pi(|0|) = []
      pi(true) = []
      pi(s) = [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

  (no rules)

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              eq#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 2
              eq#!_A(x1,x2) = x1 + x2
              s!_A(x1) = x1 + 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            eq#_A(x1,x2) = 0
            s_A(x1) = 4
            eq#!_A(x1,x2) = max{2, x1 - 1, x2 - 7}
            s!_A(x1) = max{1, x1 - 6}
        
  
    partial status:
  
      pi(eq#) = []
      pi(s) = []

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_min#_A(x1,x2) = x2
              false_A() = 15
              cons_A(x1,x2) = x1 + x2 + 7
              min#_A(x1) = x1
              le_A(x1,x2) = x1 + x2 + 8
              true_A() = 14
              |0|_A() = 6
              s_A(x1) = x1 + 1
              if_min#!_A(x1,x2) = x2 + 8
              false!_A() = 20
              cons!_A(x1,x2) = 23
              min#!_A(x1) = x1 + 10
              le!_A(x1,x2) = 22
              true!_A() = 18
              |0|!_A() = 21
              s!_A(x1) = 21
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_min#_A(x1,x2) = max{0, x1 - 217, x2 - 63}
            false_A = 1
            cons_A(x1,x2) = max{40, x1 + 24, x2 + 22}
            min#_A(x1) = max{0, x1 - 63}
            le_A(x1,x2) = max{x1 - 6, x2 + 200}
            true_A = 72
            |0|_A = 0
            s_A(x1) = x1 + 5
            if_min#!_A(x1,x2) = max{x1 - 2, x2 + 30}
            false!_A = 110
            cons!_A(x1,x2) = max{69, x1 + 48, x2 - 38}
            min#!_A(x1) = max{93, x1 + 69}
            le!_A(x1,x2) = max{3, x1 - 1, x2 - 15}
            true!_A = 87
            |0|!_A = 111
            s!_A(x1) = max{2, x1 - 16}
        
  
    partial status:
  
      pi(if_min#) = [2]
      pi(false) = []
      pi(cons) = []
      pi(min#) = []
      pi(le) = []
      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

  (no rules)

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!):
        
          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) = x1 + x2
              s!_A(x1) = 6
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le#_A(x1,x2) = 0
            s_A(x1) = 2
            le#!_A(x1,x2) = 122
            s!_A(x1) = max{121, x1 - 2}
        
  
    partial status:
  
      pi(le#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.