YES

We show the termination of the TRS R:

  and(false(),false()) -> false()
  and(true(),false()) -> false()
  and(false(),true()) -> false()
  and(true(),true()) -> true()
  eq(nil(),nil()) -> true()
  eq(cons(T,L),nil()) -> false()
  eq(nil(),cons(T,L)) -> false()
  eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
  eq(var(L),var(Lp)) -> eq(L,Lp)
  eq(var(L),apply(T,S)) -> false()
  eq(var(L),lambda(X,T)) -> false()
  eq(apply(T,S),var(L)) -> false()
  eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
  eq(apply(T,S),lambda(X,Tp)) -> false()
  eq(lambda(X,T),var(L)) -> false()
  eq(lambda(X,T),apply(Tp,Sp)) -> false()
  eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
  if(true(),var(K),var(L)) -> var(K)
  if(false(),var(K),var(L)) -> var(L)
  ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
  ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
  ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

-- SCC decomposition.

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

p1: eq#(cons(T,L),cons(Tp,Lp)) -> and#(eq(T,Tp),eq(L,Lp))
p2: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(T,Tp)
p3: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(L,Lp)
p4: eq#(var(L),var(Lp)) -> eq#(L,Lp)
p5: eq#(apply(T,S),apply(Tp,Sp)) -> and#(eq(T,Tp),eq(S,Sp))
p6: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(T,Tp)
p7: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(S,Sp)
p8: eq#(lambda(X,T),lambda(Xp,Tp)) -> and#(eq(T,Tp),eq(X,Xp))
p9: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(T,Tp)
p10: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(X,Xp)
p11: ren#(var(L),var(K),var(Lp)) -> if#(eq(L,Lp),var(K),var(Lp))
p12: ren#(var(L),var(K),var(Lp)) -> eq#(L,Lp)
p13: ren#(X,Y,apply(T,S)) -> ren#(X,Y,T)
p14: ren#(X,Y,apply(T,S)) -> ren#(X,Y,S)
p15: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))
p16: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The estimated dependency graph contains the following SCCs:

  {p13, p14, p15, p16}
  {p2, p3, p4, p6, p7, p9, p10}


-- Reduction pair.

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

p1: ren#(X,Y,lambda(Z,T)) -> ren#(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)
p2: ren#(X,Y,lambda(Z,T)) -> ren#(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T))
p3: ren#(X,Y,apply(T,S)) -> ren#(X,Y,S)
p4: ren#(X,Y,apply(T,S)) -> ren#(X,Y,T)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      ren#_A(x1,x2,x3) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x3
      lambda_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x2 + (4,1,1,2)
      var_A(x1) = (7,1,1,3)
      cons_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + x2 + (1,1,1,1)
      nil_A() = (1,1,1,1)
      ren_A(x1,x2,x3) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x3 + (0,2,0,1)
      apply_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (2,0,1,1)
      and_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x2 + (2,0,14,0)
      false_A() = (2,0,14,17)
      true_A() = (1,1,0,0)
      eq_A(x1,x2) = (3,4,1,16)
      if_A(x1,x2,x3) = (7,2,2,2)

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

p1: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(T,Tp)
p2: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(X,Xp)
p3: eq#(lambda(X,T),lambda(Xp,Tp)) -> eq#(T,Tp)
p4: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(S,Sp)
p5: eq#(apply(T,S),apply(Tp,Sp)) -> eq#(T,Tp)
p6: eq#(var(L),var(Lp)) -> eq#(L,Lp)
p7: eq#(cons(T,L),cons(Tp,Lp)) -> eq#(L,Lp)

and R consists of:

r1: and(false(),false()) -> false()
r2: and(true(),false()) -> false()
r3: and(false(),true()) -> false()
r4: and(true(),true()) -> true()
r5: eq(nil(),nil()) -> true()
r6: eq(cons(T,L),nil()) -> false()
r7: eq(nil(),cons(T,L)) -> false()
r8: eq(cons(T,L),cons(Tp,Lp)) -> and(eq(T,Tp),eq(L,Lp))
r9: eq(var(L),var(Lp)) -> eq(L,Lp)
r10: eq(var(L),apply(T,S)) -> false()
r11: eq(var(L),lambda(X,T)) -> false()
r12: eq(apply(T,S),var(L)) -> false()
r13: eq(apply(T,S),apply(Tp,Sp)) -> and(eq(T,Tp),eq(S,Sp))
r14: eq(apply(T,S),lambda(X,Tp)) -> false()
r15: eq(lambda(X,T),var(L)) -> false()
r16: eq(lambda(X,T),apply(Tp,Sp)) -> false()
r17: eq(lambda(X,T),lambda(Xp,Tp)) -> and(eq(T,Tp),eq(X,Xp))
r18: if(true(),var(K),var(L)) -> var(K)
r19: if(false(),var(K),var(L)) -> var(L)
r20: ren(var(L),var(K),var(Lp)) -> if(eq(L,Lp),var(K),var(Lp))
r21: ren(X,Y,apply(T,S)) -> apply(ren(X,Y,T),ren(X,Y,S))
r22: ren(X,Y,lambda(Z,T)) -> lambda(var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),ren(X,Y,ren(Z,var(cons(X,cons(Y,cons(lambda(Z,T),nil())))),T)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      eq#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2
      cons_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      lambda_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,0,0),(1,1,1,0)) x2 + (1,1,1,1)
      apply_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)
      var_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7

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