YES
We show the termination of the TRS R:

  min(|0|(),y) -> |0|()
  min(s(x),|0|()) -> |0|()
  min(s(x),s(y)) -> min(x,y)
  len(nil()) -> |0|()
  len(cons(x,xs)) -> s(len(xs))
  sum(x,|0|()) -> x
  sum(x,s(y)) -> s(sum(x,y))
  le(|0|(),x) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  take(|0|(),cons(y,ys)) -> y
  take(s(x),cons(y,ys)) -> take(x,ys)
  addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
  if(false(),c,x,y,z) -> z
  if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))


-- SCC decomposition.

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

p1: min#(s(x),s(y)) -> min#(x,y)
p2: len#(cons(x,xs)) -> len#(xs)
p3: sum#(x,s(y)) -> sum#(x,y)
p4: le#(s(x),s(y)) -> le#(x,y)
p5: take#(s(x),cons(y,ys)) -> take#(x,ys)
p6: addList#(x,y) -> if#(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
p7: addList#(x,y) -> le#(|0|(),min(len(x),len(y)))
p8: addList#(x,y) -> min#(len(x),len(y))
p9: addList#(x,y) -> len#(x)
p10: addList#(x,y) -> len#(y)
p11: if#(true(),c,xs,ys,z) -> if#(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))
p12: if#(true(),c,xs,ys,z) -> le#(s(c),min(len(xs),len(ys)))
p13: if#(true(),c,xs,ys,z) -> min#(len(xs),len(ys))
p14: if#(true(),c,xs,ys,z) -> len#(xs)
p15: if#(true(),c,xs,ys,z) -> len#(ys)
p16: if#(true(),c,xs,ys,z) -> sum#(take(c,xs),take(c,ys))
p17: if#(true(),c,xs,ys,z) -> take#(c,xs)
p18: if#(true(),c,xs,ys,z) -> take#(c,ys)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The estimated dependency graph contains the following SCCs:

  {p11}
  {p1}
  {p2}
  {p3}
  {p4}
  {p5}


-- Reduction pair.

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

p1: if#(true(),c,xs,ys,z) -> if#(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2, r3, r8, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if#_A(x1,x2,x3,x4,x5) = max{x1 + 13, x3 + 31, x4 + 41, x5 + 31}
        true_A = 28
        le_A(x1,x2) = max{28, x2 + 2}
        s_A(x1) = max{29, x1 + 8}
        min_A(x1,x2) = 26
        len_A(x1) = max{42, x1 - 9}
        cons_A(x1,x2) = 10
        sum_A(x1,x2) = max{42, x2 - 33}
        take_A(x1,x2) = x1 + 75
        |0|_A = 24
        false_A = 24
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if#_A(x1,x2,x3,x4,x5) = max{1, x2 - 13, x3 - 13, x4 - 13, x5 - 17}
          true_A = 84
          le_A(x1,x2) = 84
          s_A(x1) = 14
          min_A(x1,x2) = 28
          len_A(x1) = 56
          cons_A(x1,x2) = 5
          sum_A(x1,x2) = 73
          take_A(x1,x2) = max{x1 + 17, x2 + 17}
          |0|_A = 27
          false_A = 84
          if#!_A(x1,x2,x3,x4,x5) = max{x1 - 61, x2 + 8, x5 + 1}
          true!_A = 90
          le!_A(x1,x2) = max{x1 + 73, x2 + 60}
          s!_A(x1) = 0
          min!_A(x1,x2) = 0
          len!_A(x1) = 23
          cons!_A(x1,x2) = max{91, x2 - 1}
          sum!_A(x1,x2) = max{89, x2 - 15}
          take!_A(x1,x2) = max{89, x1 - 15, x2 + 17}
          |0|!_A = 24
          false!_A = 0
  
    partial status:
  
      pi(if#) = [1, 5]
      pi(true) = []
      pi(le) = [2]
      pi(s) = []
      pi(min) = []
      pi(len) = []
      pi(cons) = []
      pi(sum) = []
      pi(take) = []
      pi(|0|) = []
      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: min#(s(x),s(y)) -> min#(x,y)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),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:
      
        min#_A(x1,x2) = max{22, x1 - 1, x2 - 1}
        s_A(x1) = x1 + 48
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          min#_A(x1,x2) = 0
          s_A(x1) = max{100, x1 - 211}
          min#!_A(x1,x2) = x2 + 219
          s!_A(x1) = max{6, x1 - 552}
  
    partial status:
  
      pi(min#) = []
      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: len#(cons(x,xs)) -> len#(xs)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),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:
      
        len#_A(x1) = 1
        cons_A(x1,x2) = max{19, x1 - 15, x2 - 15}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          len#_A(x1) = x1
          cons_A(x1,x2) = max{16, x1 + 11, x2 + 11}
          len#!_A(x1) = max{3, x1 - 12}
          cons!_A(x1,x2) = max{6, x1 + 4, x2 - 309}
  
    partial status:
  
      pi(len#) = []
      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: sum#(x,s(y)) -> sum#(x,y)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sum#_A(x1,x2) = max{1, x2 - 5}
        s_A(x1) = max{26, x1 + 4}
        min_A(x1,x2) = max{27, x1 - 50, x2 - 50}
        |0|_A = 25
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sum#_A(x1,x2) = 0
          s_A(x1) = max{13, x1 - 216}
          min_A(x1,x2) = max{189, x1 - 23, x2 - 26}
          |0|_A = 22
          sum#!_A(x1,x2) = max{10, x1 - 1, x2 - 14}
          s!_A(x1) = max{9, x1 - 328}
          min!_A(x1,x2) = max{6, x1 - 218, x2 - 221}
          |0|!_A = 3
  
    partial status:
  
      pi(sum#) = []
      pi(s) = [1]
      pi(min) = []
      pi(|0|) = []

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: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),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{30, x1 + 16, x2 + 16}
        s_A(x1) = x1 + 29
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = 0
          s_A(x1) = max{68, x1 - 2}
          le#!_A(x1,x2) = x2 + 7
          s!_A(x1) = max{3, x1 - 148}
  
    partial status:
  
      pi(le#) = []
      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: take#(s(x),cons(y,ys)) -> take#(x,ys)

and R consists of:

r1: min(|0|(),y) -> |0|()
r2: min(s(x),|0|()) -> |0|()
r3: min(s(x),s(y)) -> min(x,y)
r4: len(nil()) -> |0|()
r5: len(cons(x,xs)) -> s(len(xs))
r6: sum(x,|0|()) -> x
r7: sum(x,s(y)) -> s(sum(x,y))
r8: le(|0|(),x) -> true()
r9: le(s(x),|0|()) -> false()
r10: le(s(x),s(y)) -> le(x,y)
r11: take(|0|(),cons(y,ys)) -> y
r12: take(s(x),cons(y,ys)) -> take(x,ys)
r13: addList(x,y) -> if(le(|0|(),min(len(x),len(y))),|0|(),x,y,nil())
r14: if(false(),c,x,y,z) -> z
r15: if(true(),c,xs,ys,z) -> if(le(s(c),min(len(xs),len(ys))),s(c),xs,ys,cons(sum(take(c,xs),take(c,ys)),z))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        take#_A(x1,x2) = max{21, x1 + 2}
        s_A(x1) = max{23, x1 + 3}
        cons_A(x1,x2) = max{20, x1 - 36, x2 - 36}
        min_A(x1,x2) = max{4, x1 - 24, x2 - 26}
        |0|_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          take#_A(x1,x2) = 0
          s_A(x1) = max{58, x1 - 221}
          cons_A(x1,x2) = max{65, x1 - 42, x2 - 42}
          min_A(x1,x2) = max{122, x1 - 68, x2 - 98}
          |0|_A = 67
          take#!_A(x1,x2) = max{0, x1 - 59}
          s!_A(x1) = 0
          cons!_A(x1,x2) = max{5, x1 - 121, x2 - 76}
          min!_A(x1,x2) = max{3, x1 - 294, x2 - 687}
          |0|!_A = 2
  
    partial status:
  
      pi(take#) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(min) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.