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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,1)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,7)
            forall_A() = (4,2)
            cons_A() = (1,0)
            forsome_A() = (4,0)
    
      precedence:
      
        cons > app# = app = forall = forsome
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(forall) = []
        pi(cons) = []
        pi(forsome) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (0,0)
            app_A(x1,x2) = (3,3)
            forall_A() = (1,1)
            cons_A() = (2,2)
            forsome_A() = (6,4)
    
      precedence:
      
        app = forsome > app# = forall = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(forall) = []
        pi(cons) = []
        pi(forsome) = []
    

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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)

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:

  {p1, p2, p3}


-- 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#(p,x)
p3: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + (3,4)
            app_A(x1,x2) = x2 + (0,3)
            forall_A() = (1,1)
            cons_A() = (2,6)
            forsome_A() = (1,5)
    
      precedence:
      
        forall = cons > app# = forsome > app
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = []
        pi(forall) = []
        pi(cons) = []
        pi(forsome) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (0,0)
            app_A(x1,x2) = (1,0)
            forall_A() = (1,1)
            cons_A() = (3,2)
            forsome_A() = (2,1)
    
      precedence:
      
        forsome > app > app# = forall = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(forall) = []
        pi(cons) = []
        pi(forsome) = []
    

The next rules are strictly ordered:

  p1

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#(p,x)
p2: 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:

  {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#(p,x)
p2: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x2 + (5,1)
            app_A(x1,x2) = x1 + x2 + (2,2)
            forsome_A() = (1,2)
            cons_A() = (2,3)
    
      precedence:
      
        app# = app = forsome = cons
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = [1, 2]
        pi(forsome) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x2
            app_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,3)
            forsome_A() = (1,1)
            cons_A() = (2,2)
    
      precedence:
      
        app# = app = forsome > cons
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = []
        pi(forsome) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1, p2

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