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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        a__and#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,1)) x2 + (2,10,15,5)
        true_A() = (1,15,0,2)
        mark#_A(x1) = x1 + (0,16,14,6)
        first_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x2 + (1,0,0,0)
        add_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,1)) x1 + x2 + (7,0,3,1)
        a__add#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (1,0,18,0)
        mark_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,0,0,1)) x1 + (0,5,8,0)
        |0|_A() = (1,1,0,0)
        if_A(x1,x2,x3) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x2 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x3 + (1,1,0,0)
        a__if#_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,0,0)) x1 + x2 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x3
        false_A() = (1,0,15,0)
        and_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,1,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x2 + (2,0,10,0)
        a__and_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,1)) x2 + (2,0,11,0)
        a__if_A(x1,x2,x3) = x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + x3 + (1,4,1,1)
        a__add_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,1)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (7,0,4,2)
        s_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + (1,0,0,0)
        a__first_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + x2 + (1,4,3,0)
        nil_A() = (0,0,0,1)
        cons_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x2 + (0,3,1,1)
        a__from_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,2,10,3)
        from_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,1,0,4)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        a__add > false > add > true > mark > a__from > from > a__and > a__if > if > and > a__first > first > |0| > mark# > a__add# > cons > nil > s > a__and# > a__if#
      
      argument filter:
    
        pi(a__and#) = 2
        pi(true) = []
        pi(mark#) = 1
        pi(first) = []
        pi(add) = 1
        pi(a__add#) = []
        pi(mark) = [1]
        pi(|0|) = []
        pi(if) = []
        pi(a__if#) = []
        pi(false) = []
        pi(and) = 2
        pi(a__and) = [2]
        pi(a__if) = []
        pi(a__add) = 1
        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.