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. lexicographic path order with precedence:
    
      precedence:
      
        forsome > app# > forall > app > cons
      
      argument filter:
    
        pi(app#) = 1
        pi(app) = 2
        pi(forall) = []
        pi(cons) = []
        pi(forsome) = []
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        app#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1
        app_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
        forall_A() = (1,1,1)
        cons_A() = (1,1,0)
        forsome_A() = (1,1,1)
    

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(),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#(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. lexicographic path order with precedence:
    
      precedence:
      
        app# > cons > app > forsome
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(forsome) = []
        pi(cons) = []
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        app#_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x2
        app_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,0)) x1 + ((1,0,0),(1,1,0),(0,0,0)) x2
        forsome_A() = (0,0,0)
        cons_A() = (1,1,0)
    

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(),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. lexicographic path order with precedence:
    
      precedence:
      
        app# > cons > app > forall
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(forall) = []
        pi(cons) = []
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        app#_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x2
        app_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,0)) x1 + ((1,0,0),(1,1,0),(0,0,0)) x2
        forall_A() = (0,0,0)
        cons_A() = (1,1,0)
    

The next rules are strictly ordered:

  p1

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