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))

-- SCC decomposition.

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

p1: app#(app(forall(),p),app(app(cons(),x),xs)) -> app#(app(and(),app(p,x)),app(app(forall(),p),xs))
p2: app#(app(forall(),p),app(app(cons(),x),xs)) -> app#(and(),app(p,x))
p3: app#(app(forall(),p),app(app(cons(),x),xs)) -> app#(p,x)
p4: app#(app(forall(),p),app(app(cons(),x),xs)) -> app#(app(forall(),p),xs)
p5: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(app(or(),app(p,x)),app(app(forsome(),p),xs))
p6: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(or(),app(p,x))
p7: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(p,x)
p8: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(app(forsome(),p),xs)

and R consists of:

r1: app(app(and(),true()),true()) -> true()
r2: app(app(and(),true()),false()) -> false()
r3: app(app(and(),false()),true()) -> false()
r4: app(app(and(),false()),false()) -> false()
r5: app(app(or(),true()),true()) -> true()
r6: app(app(or(),true()),false()) -> true()
r7: app(app(or(),false()),true()) -> true()
r8: app(app(or(),false()),false()) -> false()
r9: app(app(forall(),p),nil()) -> true()
r10: app(app(forall(),p),app(app(cons(),x),xs)) -> app(app(and(),app(p,x)),app(app(forall(),p),xs))
r11: app(app(forsome(),p),nil()) -> false()
r12: app(app(forsome(),p),app(app(cons(),x),xs)) -> app(app(or(),app(p,x)),app(app(forsome(),p),xs))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p7, p8}


-- Reduction pair.

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

p1: app#(app(forall(),p),app(app(cons(),x),xs)) -> app#(p,x)
p2: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(app(forsome(),p),xs)
p3: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(p,x)
p4: app#(app(forall(),p),app(app(cons(),x),xs)) -> app#(app(forall(),p),xs)

and R consists of:

r1: app(app(and(),true()),true()) -> true()
r2: app(app(and(),true()),false()) -> false()
r3: app(app(and(),false()),true()) -> false()
r4: app(app(and(),false()),false()) -> false()
r5: app(app(or(),true()),true()) -> true()
r6: app(app(or(),true()),false()) -> true()
r7: app(app(or(),false()),true()) -> true()
r8: app(app(or(),false()),false()) -> false()
r9: app(app(forall(),p),nil()) -> true()
r10: app(app(forall(),p),app(app(cons(),x),xs)) -> app(app(and(),app(p,x)),app(app(forall(),p),xs))
r11: app(app(forsome(),p),nil()) -> false()
r12: app(app(forsome(),p),app(app(cons(),x),xs)) -> app(app(or(),app(p,x)),app(app(forsome(),p),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,1,0)) x2
      app_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,1,1)) x2 + (1,1,0,0)
      forall_A() = (1,1,1,1)
      cons_A() = (0,1,1,1)
      forsome_A() = (1,1,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4

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