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 path order with precedence:
  
    precedence:
    
      app > forsome > forall > app# > cons
    
    argument filter:
  
      pi(app#) = [1]
      pi(app) = [1, 2]
      pi(forall) = []
      pi(cons) = []
      pi(forsome) = []

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(app(forsome(),p),xs)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1}
  {p2}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      app > forsome > app# > cons
    
    argument filter:
  
      pi(app#) = [1, 2]
      pi(app) = [1, 2]
      pi(forsome) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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: 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 monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      app > forall > app# > cons
    
    argument filter:
  
      pi(app#) = [1, 2]
      pi(app) = [1, 2]
      pi(forall) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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