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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 1, x2 - 5}
          app_A(x1,x2) = max{14, x1 + 3, x2 + 11}
          forall_A = 13
          cons_A = 9
          forsome_A = 1
    
      precedence:
      
        app# = app = forall = cons = forsome
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = [1, 2]
        pi(forall) = []
        pi(cons) = []
        pi(forsome) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = x1 + 4
          app_A(x1,x2) = max{x1 + 8, x2 + 10}
          forall_A = 7
          cons_A = 8
          forsome_A = 9
    
      precedence:
      
        app# = app = forall = cons = forsome
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = []
        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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x2 - 4}
          app_A(x1,x2) = max{x1 + 6, x2 + 3}
          forsome_A = 2
          cons_A = 3
    
      precedence:
      
        app# = app = forsome = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(forsome) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = 0
          app_A(x1,x2) = max{x1 + 2, x2 + 6}
          forsome_A = 3
          cons_A = 4
    
      precedence:
      
        app# = app = forsome = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(forsome) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{0, x2 - 5}
          app_A(x1,x2) = max{x1 + 7, x2 + 4}
          forall_A = 3
          cons_A = 4
    
      precedence:
      
        app# = app = forall = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(forall) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = 0
          app_A(x1,x2) = max{x1 + 2, x2 + 6}
          forall_A = 3
          cons_A = 4
    
      precedence:
      
        app# = app = forall = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(forall) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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