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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(1,0)) x2 + (5,4)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,2)
          forall_A() = (1,1)
          cons_A() = (2,3)
          forsome_A() = (10,5)
  
    precedence:
    
      app# > app = forall = cons = forsome
    
    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#(app(forsome(),p),xs)
p2: app#(app(forsome(),p),app(app(cons(),x),xs)) -> app#(p,x)
p3: 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}
  {p3}


-- 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)
p2: 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(),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:

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

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)

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}


-- 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)

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:

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (4,4)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,3)
          forall_A() = (1,1)
          cons_A() = (3,2)
  
    precedence:
    
      app# = app = forall = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = []
      pi(forall) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

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