YES
We show the termination of the TRS R:

  empty(nil()) -> true()
  empty(cons(x,l)) -> false()
  head(cons(x,l)) -> x
  tail(nil()) -> nil()
  tail(cons(x,l)) -> l
  rev(nil()) -> nil()
  rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
  last(x,l) -> if(empty(l),x,l)
  if(true(),x,l) -> x
  if(false(),x,l) -> last(head(l),tail(l))
  rev2(x,nil()) -> nil()
  rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))


-- SCC decomposition.

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

p1: rev#(cons(x,l)) -> rev2#(x,l)
p2: last#(x,l) -> if#(empty(l),x,l)
p3: last#(x,l) -> empty#(l)
p4: if#(false(),x,l) -> last#(head(l),tail(l))
p5: if#(false(),x,l) -> head#(l)
p6: if#(false(),x,l) -> tail#(l)
p7: rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))
p8: rev2#(x,cons(y,l)) -> rev2#(y,l)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,l)) -> false()
r3: head(cons(x,l)) -> x
r4: tail(nil()) -> nil()
r5: tail(cons(x,l)) -> l
r6: rev(nil()) -> nil()
r7: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r8: last(x,l) -> if(empty(l),x,l)
r9: if(true(),x,l) -> x
r10: if(false(),x,l) -> last(head(l),tail(l))
r11: rev2(x,nil()) -> nil()
r12: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

The estimated dependency graph contains the following SCCs:

  {p1, p7, p8}
  {p2, p4}


-- Reduction pair.

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

p1: rev#(cons(x,l)) -> rev2#(x,l)
p2: rev2#(x,cons(y,l)) -> rev2#(y,l)
p3: rev2#(x,cons(y,l)) -> rev#(cons(x,rev2(y,l)))

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,l)) -> false()
r3: head(cons(x,l)) -> x
r4: tail(nil()) -> nil()
r5: tail(cons(x,l)) -> l
r6: rev(nil()) -> nil()
r7: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r8: last(x,l) -> if(empty(l),x,l)
r9: if(true(),x,l) -> x
r10: if(false(),x,l) -> last(head(l),tail(l))
r11: rev2(x,nil()) -> nil()
r12: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

The set of usable rules consists of

  r4, r6, r7, r11, r12

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        rev#_A(x1) = max{15, x1 + 3}
        cons_A(x1,x2) = max{21, x2 + 13}
        rev2#_A(x1,x2) = max{14, x2 + 12}
        rev2_A(x1,x2) = max{2, x2}
        tail_A(x1) = max{17, x1 - 14}
        nil_A = 1
        rev_A(x1) = max{16, x1}
        rev1_A(x1,x2) = max{13, x1 - 1, x2 + 1}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          rev#_A(x1) = 1
          cons_A(x1,x2) = x1 + 9
          rev2#_A(x1,x2) = 1
          rev2_A(x1,x2) = x1 + 12
          tail_A(x1) = max{16, x1 - 10}
          nil_A = 9
          rev_A(x1) = max{8, x1 + 3}
          rev1_A(x1,x2) = max{3, x1}
          rev#!_A(x1) = max{10, x1 + 4}
          cons!_A(x1,x2) = max{11, x2 + 9}
          rev2#!_A(x1,x2) = max{14, x1 + 11, x2 + 9}
          rev2!_A(x1,x2) = max{x1 + 23, x2 - 10}
          tail!_A(x1) = max{25, x1 - 10}
          nil!_A = 24
          rev!_A(x1) = x1 + 13
          rev1!_A(x1,x2) = max{23, x1 - 1}
  
    partial status:
  
      pi(rev#) = [1]
      pi(cons) = []
      pi(rev2#) = []
      pi(rev2) = []
      pi(tail) = []
      pi(nil) = []
      pi(rev) = []
      pi(rev1) = [2]

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: last#(x,l) -> if#(empty(l),x,l)
p2: if#(false(),x,l) -> last#(head(l),tail(l))

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,l)) -> false()
r3: head(cons(x,l)) -> x
r4: tail(nil()) -> nil()
r5: tail(cons(x,l)) -> l
r6: rev(nil()) -> nil()
r7: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l))
r8: last(x,l) -> if(empty(l),x,l)
r9: if(true(),x,l) -> x
r10: if(false(),x,l) -> last(head(l),tail(l))
r11: rev2(x,nil()) -> nil()
r12: rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l)))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        last#_A(x1,x2) = x2 + 5
        if#_A(x1,x2,x3) = max{x1 + 2, x3 + 4}
        empty_A(x1) = max{2, x1 - 10}
        false_A = 12
        head_A(x1) = max{26, x1 + 24}
        tail_A(x1) = max{8, x1 - 6}
        nil_A = 7
        true_A = 0
        cons_A(x1,x2) = max{x1 - 23, x2 + 22}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          last#_A(x1,x2) = max{0, x1 - 14, x2 - 12}
          if#_A(x1,x2,x3) = max{0, x1 - 12, x2 - 14, x3 - 12}
          empty_A(x1) = max{1, x1 - 9}
          false_A = 0
          head_A(x1) = max{3, x1 + 2}
          tail_A(x1) = max{11, x1 - 4}
          nil_A = 3
          true_A = 0
          cons_A(x1,x2) = max{x1 - 2, x2 + 8}
          last#!_A(x1,x2) = max{7, x1 - 4, x2 - 17}
          if#!_A(x1,x2,x3) = max{6, x1 - 9, x2 - 5}
          empty!_A(x1) = max{2, x1 - 18}
          false!_A = 1
          head!_A(x1) = max{5, x1 - 3}
          tail!_A(x1) = max{4, x1 - 4}
          nil!_A = 1
          true!_A = 0
          cons!_A(x1,x2) = max{3, x1 - 23, x2 - 6}
  
    partial status:
  
      pi(last#) = []
      pi(if#) = [3]
      pi(empty) = []
      pi(false) = []
      pi(head) = []
      pi(tail) = []
      pi(nil) = []
      pi(true) = []
      pi(cons) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.