YES

We show the termination of the TRS R:

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

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        app_A(x1,x2) = x1 + x2
        forall_A() = 1
        cons_A() = 1
        forsome_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        app#_A(x1,x2) = x1 + x2
        app_A(x1,x2) = x2 + 1
        forall_A() = 1
        cons_A() = 0
        forsome_A() = 1
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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