YES
We show the termination of the TRS R:

  app(app(and(),true()),true()) -> true()
  app(app(and(),true()),false()) -> false()
  app(app(and(),false()),true()) -> false()
  app(app(and(),false()),false()) -> false()
  app(app(or(),true()),true()) -> true()
  app(app(or(),true()),false()) -> true()
  app(app(or(),false()),true()) -> true()
  app(app(or(),false()),false()) -> false()
  app(app(forall(),p),nil()) -> true()
  app(app(forall(),p),app(app(cons(),x),xs)) -> app(app(and(),app(p,x)),app(app(forall(),p),xs))
  app(app(forsome(),p),nil()) -> false()
  app(app(forsome(),p),app(app(cons(),x),xs)) -> app(app(or(),app(p,x)),app(app(forsome(),p),xs))

Instead we prove termination of the uncurried system:

  app(and_0(),y) -> and_1(y)
  app(and_1(x1),y) -> and_2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(forall_0(),y) -> forall_1(y)
  app(forall_1(x1),y) -> forall_2(x1,y)
  app(forsome_0(),y) -> forsome_1(y)
  app(forsome_1(x1),y) -> forsome_2(x1,y)
  app(or_0(),y) -> or_1(y)
  app(or_1(x1),y) -> or_2(x1,y)
  and_2(true_0(),true_0()) -> true_0()
  and_2(true_0(),false_0()) -> false_0()
  and_2(false_0(),true_0()) -> false_0()
  and_2(false_0(),false_0()) -> false_0()
  or_2(true_0(),true_0()) -> true_0()
  or_2(true_0(),false_0()) -> true_0()
  or_2(false_0(),true_0()) -> true_0()
  or_2(false_0(),false_0()) -> false_0()
  forall_2(p,nil_0()) -> true_0()
  forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs))
  forsome_2(p,nil_0()) -> false_0()
  forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs))


-- SCC decomposition.

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

p1: app#(and_1(x1),y) -> and_2#(x1,y)
p2: app#(forall_1(x1),y) -> forall_2#(x1,y)
p3: app#(forsome_1(x1),y) -> forsome_2#(x1,y)
p4: app#(or_1(x1),y) -> or_2#(x1,y)
p5: forall_2#(p,cons_2(x,xs)) -> and_2#(app(p,x),forall_2(p,xs))
p6: forall_2#(p,cons_2(x,xs)) -> app#(p,x)
p7: forall_2#(p,cons_2(x,xs)) -> forall_2#(p,xs)
p8: forsome_2#(p,cons_2(x,xs)) -> or_2#(app(p,x),forsome_2(p,xs))
p9: forsome_2#(p,cons_2(x,xs)) -> app#(p,x)
p10: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(forall_0(),y) -> forall_1(y)
r6: app(forall_1(x1),y) -> forall_2(x1,y)
r7: app(forsome_0(),y) -> forsome_1(y)
r8: app(forsome_1(x1),y) -> forsome_2(x1,y)
r9: app(or_0(),y) -> or_1(y)
r10: app(or_1(x1),y) -> or_2(x1,y)
r11: and_2(true_0(),true_0()) -> true_0()
r12: and_2(true_0(),false_0()) -> false_0()
r13: and_2(false_0(),true_0()) -> false_0()
r14: and_2(false_0(),false_0()) -> false_0()
r15: or_2(true_0(),true_0()) -> true_0()
r16: or_2(true_0(),false_0()) -> true_0()
r17: or_2(false_0(),true_0()) -> true_0()
r18: or_2(false_0(),false_0()) -> false_0()
r19: forall_2(p,nil_0()) -> true_0()
r20: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs))
r21: forsome_2(p,nil_0()) -> false_0()
r22: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p6, p7, p9, p10}


-- Reduction pair.

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

p1: app#(forall_1(x1),y) -> forall_2#(x1,y)
p2: forall_2#(p,cons_2(x,xs)) -> forall_2#(p,xs)
p3: forall_2#(p,cons_2(x,xs)) -> app#(p,x)
p4: app#(forsome_1(x1),y) -> forsome_2#(x1,y)
p5: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs)
p6: forsome_2#(p,cons_2(x,xs)) -> app#(p,x)

and R consists of:

r1: app(and_0(),y) -> and_1(y)
r2: app(and_1(x1),y) -> and_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(forall_0(),y) -> forall_1(y)
r6: app(forall_1(x1),y) -> forall_2(x1,y)
r7: app(forsome_0(),y) -> forsome_1(y)
r8: app(forsome_1(x1),y) -> forsome_2(x1,y)
r9: app(or_0(),y) -> or_1(y)
r10: app(or_1(x1),y) -> or_2(x1,y)
r11: and_2(true_0(),true_0()) -> true_0()
r12: and_2(true_0(),false_0()) -> false_0()
r13: and_2(false_0(),true_0()) -> false_0()
r14: and_2(false_0(),false_0()) -> false_0()
r15: or_2(true_0(),true_0()) -> true_0()
r16: or_2(true_0(),false_0()) -> true_0()
r17: or_2(false_0(),true_0()) -> true_0()
r18: or_2(false_0(),false_0()) -> false_0()
r19: forall_2(p,nil_0()) -> true_0()
r20: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs))
r21: forsome_2(p,nil_0()) -> false_0()
r22: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 26, x2 - 8}
            forall_1_A(x1) = max{20, x1 + 13}
            forall_2#_A(x1,x2) = max{0, x1 - 26, x2 - 8}
            cons_2_A(x1,x2) = max{7, x1 + 2, x2 + 1}
            forsome_1_A(x1) = x1 + 25
            forsome_2#_A(x1,x2) = max{0, x1 - 26, x2 - 8}
            app#!_A(x1,x2) = max{4, x1 - 14, x2 - 21}
            forall_1!_A(x1) = max{6, x1 - 19}
            forall_2#!_A(x1,x2) = max{5, x1 - 13, x2 - 23}
            cons_2!_A(x1,x2) = max{3, x2 - 21}
            forsome_1!_A(x1) = max{6, x1 - 24}
            forsome_2#!_A(x1,x2) = max{5, x1 - 1, x2 - 22}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 7, x2 - 7}
            forall_1_A(x1) = max{6, x1 + 2}
            forall_2#_A(x1,x2) = max{0, x1 - 7, x2 - 7}
            cons_2_A(x1,x2) = max{6, x1, x2}
            forsome_1_A(x1) = x1 + 6
            forsome_2#_A(x1,x2) = max{0, x1 - 1, x2 - 7}
            app#!_A(x1,x2) = max{1, x1 - 25}
            forall_1!_A(x1) = max{4, x1 - 25}
            forall_2#!_A(x1,x2) = max{2, x1 - 24, x2 - 7}
            cons_2!_A(x1,x2) = max{0, x1 - 25, x2 - 36}
            forsome_1!_A(x1) = max{1, x1 - 23}
            forsome_2#!_A(x1,x2) = max{2, x1 - 24, x2 - 7}
        
  
    partial status:
  
      pi(app#) = [2]
      pi(forall_1) = [1]
      pi(forall_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(forsome_1) = [1]
      pi(forsome_2#) = [2]

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.