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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a__and#_A(x1,x2) = x2 + (2,3)
      true_A() = (1,1)
      mark#_A(x1) = ((1,0),(1,1)) x1
      first_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,1)
      add_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
      a__add#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,2)
      mark_A(x1) = ((1,0),(1,1)) x1 + (0,1)
      |0|_A() = (0,0)
      if_A(x1,x2,x3) = x1 + ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (2,1)
      a__if#_A(x1,x2,x3) = ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (1,2)
      false_A() = (1,1)
      and_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
      a__and_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,2)
      a__if_A(x1,x2,x3) = x1 + ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (2,2)
      a__add_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,3)
      s_A(x1) = (0,1)
      a__first_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,2)
      nil_A() = (1,3)
      cons_A(x1,x2) = x1 + (1,2)
      a__from_A(x1) = x1 + (2,2)
      from_A(x1) = x1 + (2,1)

The next rules are strictly ordered:

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

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(and(X1,X2)) -> a__and#(mark(X1),X2)

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:

  (no SCCs)