YES

We show the termination of the TRS R:

  a__and(true(),X) -> mark(X)
  a__and(false(),Y) -> false()
  a__if(true(),X,Y) -> mark(X)
  a__if(false(),X,Y) -> mark(Y)
  a__add(|0|(),X) -> mark(X)
  a__add(s(X),Y) -> s(add(X,Y))
  a__first(|0|(),X) -> nil()
  a__first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
  a__from(X) -> cons(X,from(s(X)))
  mark(and(X1,X2)) -> a__and(mark(X1),X2)
  mark(if(X1,X2,X3)) -> a__if(mark(X1),X2,X3)
  mark(add(X1,X2)) -> a__add(mark(X1),X2)
  mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
  mark(from(X)) -> a__from(X)
  mark(true()) -> true()
  mark(false()) -> false()
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(X)
  mark(nil()) -> nil()
  mark(cons(X1,X2)) -> cons(X1,X2)
  a__and(X1,X2) -> and(X1,X2)
  a__if(X1,X2,X3) -> if(X1,X2,X3)
  a__add(X1,X2) -> add(X1,X2)
  a__first(X1,X2) -> first(X1,X2)
  a__from(X) -> from(X)

-- SCC decomposition.

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

p1: a__and#(true(),X) -> mark#(X)
p2: a__if#(true(),X,Y) -> mark#(X)
p3: a__if#(false(),X,Y) -> mark#(Y)
p4: a__add#(|0|(),X) -> mark#(X)
p5: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p6: mark#(and(X1,X2)) -> mark#(X1)
p7: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),X2,X3)
p8: mark#(if(X1,X2,X3)) -> mark#(X1)
p9: mark#(add(X1,X2)) -> a__add#(mark(X1),X2)
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p12: mark#(first(X1,X2)) -> mark#(X1)
p13: mark#(first(X1,X2)) -> mark#(X2)
p14: mark#(from(X)) -> a__from#(X)

and R consists of:

r1: a__and(true(),X) -> mark(X)
r2: a__and(false(),Y) -> false()
r3: a__if(true(),X,Y) -> mark(X)
r4: a__if(false(),X,Y) -> mark(Y)
r5: a__add(|0|(),X) -> mark(X)
r6: a__add(s(X),Y) -> s(add(X,Y))
r7: a__first(|0|(),X) -> nil()
r8: a__first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
r9: a__from(X) -> cons(X,from(s(X)))
r10: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r11: mark(if(X1,X2,X3)) -> a__if(mark(X1),X2,X3)
r12: mark(add(X1,X2)) -> a__add(mark(X1),X2)
r13: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r14: mark(from(X)) -> a__from(X)
r15: mark(true()) -> true()
r16: mark(false()) -> false()
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(X)
r19: mark(nil()) -> nil()
r20: mark(cons(X1,X2)) -> cons(X1,X2)
r21: a__and(X1,X2) -> and(X1,X2)
r22: a__if(X1,X2,X3) -> if(X1,X2,X3)
r23: a__add(X1,X2) -> add(X1,X2)
r24: a__first(X1,X2) -> first(X1,X2)
r25: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p12, p13}


-- Reduction pair.

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

p1: a__and#(true(),X) -> mark#(X)
p2: mark#(first(X1,X2)) -> mark#(X2)
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> a__add#(mark(X1),X2)
p6: a__add#(|0|(),X) -> mark#(X)
p7: mark#(if(X1,X2,X3)) -> mark#(X1)
p8: mark#(if(X1,X2,X3)) -> a__if#(mark(X1),X2,X3)
p9: a__if#(false(),X,Y) -> mark#(Y)
p10: mark#(and(X1,X2)) -> mark#(X1)
p11: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)
p12: a__if#(true(),X,Y) -> mark#(X)

and R consists of:

r1: a__and(true(),X) -> mark(X)
r2: a__and(false(),Y) -> false()
r3: a__if(true(),X,Y) -> mark(X)
r4: a__if(false(),X,Y) -> mark(Y)
r5: a__add(|0|(),X) -> mark(X)
r6: a__add(s(X),Y) -> s(add(X,Y))
r7: a__first(|0|(),X) -> nil()
r8: a__first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
r9: a__from(X) -> cons(X,from(s(X)))
r10: mark(and(X1,X2)) -> a__and(mark(X1),X2)
r11: mark(if(X1,X2,X3)) -> a__if(mark(X1),X2,X3)
r12: mark(add(X1,X2)) -> a__add(mark(X1),X2)
r13: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r14: mark(from(X)) -> a__from(X)
r15: mark(true()) -> true()
r16: mark(false()) -> false()
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(X)
r19: mark(nil()) -> nil()
r20: mark(cons(X1,X2)) -> cons(X1,X2)
r21: a__and(X1,X2) -> and(X1,X2)
r22: a__if(X1,X2,X3) -> if(X1,X2,X3)
r23: a__add(X1,X2) -> add(X1,X2)
r24: a__first(X1,X2) -> first(X1,X2)
r25: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__and#_A(x1,x2) = max{29, x1 + 15, x2 + 9}
          true_A = 13
          mark#_A(x1) = x1 + 8
          first_A(x1,x2) = max{33, x1 + 7, x2 + 20}
          add_A(x1,x2) = max{x1 + 3, x2 + 29}
          a__add#_A(x1,x2) = max{x1 + 3, x2 + 29}
          mark_A(x1) = max{12, x1}
          |0|_A = 13
          if_A(x1,x2,x3) = max{x1 + 7, x2 + 20, x3 + 20}
          a__if#_A(x1,x2,x3) = max{x1 + 1, x2 + 28, x3 + 19}
          false_A = 13
          and_A(x1,x2) = max{x1 + 15, x2 + 29}
          a__and_A(x1,x2) = max{x1 + 15, x2 + 29}
          a__if_A(x1,x2,x3) = max{x1 + 7, x2 + 20, x3 + 20}
          a__add_A(x1,x2) = max{x1 + 3, x2 + 29}
          s_A(x1) = max{28, x1 - 2}
          a__first_A(x1,x2) = max{33, x1 + 7, x2 + 20}
          nil_A = 1
          cons_A(x1,x2) = x1 + 4
          a__from_A(x1) = x1 + 4
          from_A(x1) = x1 + 4
    
      precedence:
      
        nil > mark = false = and = a__and = a__if = a__first = a__from = from > first > a__add = cons > a__and# = a__add# = s > true = mark# = add = |0| > a__if# > if
      
      partial status:
    
        pi(a__and#) = [1, 2]
        pi(true) = []
        pi(mark#) = [1]
        pi(first) = []
        pi(add) = [1, 2]
        pi(a__add#) = [2]
        pi(mark) = [1]
        pi(|0|) = []
        pi(if) = [1]
        pi(a__if#) = [1, 2, 3]
        pi(false) = []
        pi(and) = [1, 2]
        pi(a__and) = [1, 2]
        pi(a__if) = [1, 3]
        pi(a__add) = []
        pi(s) = []
        pi(a__first) = []
        pi(nil) = []
        pi(cons) = [1]
        pi(a__from) = [1]
        pi(from) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__and#_A(x1,x2) = max{x1 - 7, x2}
          true_A = 31
          mark#_A(x1) = max{23, x1}
          first_A(x1,x2) = 24
          add_A(x1,x2) = max{x1 + 25, x2 + 25}
          a__add#_A(x1,x2) = x2 + 24
          mark_A(x1) = 6
          |0|_A = 7
          if_A(x1,x2,x3) = x1 + 5
          a__if#_A(x1,x2,x3) = max{x2 + 24, x3 + 25}
          false_A = 35
          and_A(x1,x2) = max{24, x1 + 1, x2}
          a__and_A(x1,x2) = max{x1 + 28, x2 + 36}
          a__if_A(x1,x2,x3) = max{6, x1}
          a__add_A(x1,x2) = 50
          s_A(x1) = 49
          a__first_A(x1,x2) = 5
          nil_A = 8
          cons_A(x1,x2) = 6
          a__from_A(x1) = 6
          from_A(x1) = 17
    
      precedence:
      
        true > and > a__and# > mark# = mark = |0| = a__if# = a__add = nil > first = add = a__add# = if = a__from = from > false = a__and = a__if = s = a__first = cons
      
      partial status:
    
        pi(a__and#) = [2]
        pi(true) = []
        pi(mark#) = [1]
        pi(first) = []
        pi(add) = []
        pi(a__add#) = [2]
        pi(mark) = []
        pi(|0|) = []
        pi(if) = []
        pi(a__if#) = [2, 3]
        pi(false) = []
        pi(and) = [1, 2]
        pi(a__and) = []
        pi(a__if) = []
        pi(a__add) = []
        pi(s) = []
        pi(a__first) = []
        pi(nil) = []
        pi(cons) = []
        pi(a__from) = []
        pi(from) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12

We remove them from the problem.  Then no dependency pair remains.