YES

Problem:
 not(T()) -> F()
 not(F()) -> T()
 or(x,T()) -> T()
 or(x,F()) -> x
 or(x,y) -> or(y,x)
 or(or(x,y),z) -> or(x,or(y,z))
 and(x,T()) -> x
 and(x,F()) -> F()
 and(x,y) -> and(y,x)
 and(and(x,y),z) -> and(x,and(y,z))
 imply(x,y) -> or(not(x),y)

Proof:
 sorted: (order)
 0:not(T()) -> F()
   not(F()) -> T()
   or(x,T()) -> T()
   or(x,F()) -> x
   or(x,y) -> or(y,x)
   or(or(x,y),z) -> or(x,or(y,z))
   imply(x,y) -> or(not(x),y)
 1:and(x,T()) -> x
   and(x,F()) -> F()
   and(x,y) -> and(y,x)
   and(and(x,y),z) -> and(x,and(y,z))
 -----
 sorts
   [0>2, 1>7, 1>8, 2>3, 2>4, 4>5, 5>6, 5>7, 5>8]
 sort attachment (non-strict)
   not : 5 -> 4
   T : 8
   F : 7
   or : 2 x 2 -> 2
   and : 1 x 1 -> 1
   imply : 6 x 3 -> 0
 -----
 0:not(T()) -> F()
   not(F()) -> T()
   or(x,T()) -> T()
   or(x,F()) -> x
   or(x,y) -> or(y,x)
   or(or(x,y),z) -> or(x,or(y,z))
   imply(x,y) -> or(not(x),y)
   AT confluence processor
    Complete TRS T' of input TRS:
    not(T()) -> F()
    not(F()) -> T()
    or(x,T()) -> T()
    or(x,F()) -> x
    imply(x,y) -> or(not(x),y)
    or(T(),x) -> T()
    or(F(),x) -> x
    or(x,y) -> or(y,x)
    or(or(x,y),z) -> or(x,or(y,z))
    
     T' = (P union S) with
    
     TRS P:or(x,y) -> or(y,x)
           or(or(x,y),z) -> or(x,or(y,z))
    
     TRS S:not(T()) -> F()
           not(F()) -> T()
           or(x,T()) -> T()
           or(x,F()) -> x
           imply(x,y) -> or(not(x),y)
           or(T(),x) -> T()
           or(F(),x) -> x
    
    S is linear and P is reversible.
    
     CP(S,S) = 
    T() = T(), F() = F()
    
     CP(S,P union P^-1) = 
    T() = or(T(),x), T() = or(x,or(y,T())), or(T(),z) = or(x,or(T(),z)), 
    T() = or(T(),y), or(x,T()) = or(or(x,y),T()), x = or(F(),x), or(x,y) = 
    or(x,or(y,F())), or(x,z) = or(x,or(F(),z)), y = or(F(),y), or(x,y) = 
    or(or(x,y),F()), T() = or(y,T()), or(T(),z) = or(T(),or(y,z)), T() = 
    or(x,T()), T() = or(or(T(),y),z), or(x,T()) = or(or(x,T()),z), y = 
    or(y,F()), or(y,z) = or(F(),or(y,z)), x = or(x,F()), or(y,z) = or(or(F(),y),z), 
    or(x,z) = or(or(x,F()),z)
    
     CP(P union P^-1,S) = 
    or(T(),x) = T(), or(F(),x) = x, or(x,T()) = T(), or(x,F()) = x, or(x853,or(x854,T())) = 
    T(), or(x856,or(x857,F())) = or(x856,x857), or(or(T(),x868),x869) = 
    T(), or(or(F(),x871),x872) = or(x871,x872)
    
    
    We have to check termination of S:
    
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [not](x0) = x0 + 2,
      
      [or](x0, x1) = 2x0 + x1,
      
      [imply](x0, x1) = 2x0 + x1 + 4,
      
      [T] = 3,
      
      [F] = 1
     orientation:
      not(T()) = 5 >= 1 = F()
      
      not(F()) = 3 >= 3 = T()
      
      or(x,T()) = 2x + 3 >= 3 = T()
      
      or(x,F()) = 2x + 1 >= x = x
      
      imply(x,y) = 2x + y + 4 >= 2x + y + 4 = or(not(x),y)
      
      or(T(),x) = x + 6 >= 3 = T()
      
      or(F(),x) = x + 2 >= x = x
     problem:
      not(F()) -> T()
      or(x,T()) -> T()
      imply(x,y) -> or(not(x),y)
     Matrix Interpretation Processor: dim=1
      
      interpretation:
       [not](x0) = x0,
       
       [or](x0, x1) = 2x0 + x1,
       
       [imply](x0, x1) = 2x0 + x1,
       
       [T] = 0,
       
       [F] = 4
      orientation:
       not(F()) = 4 >= 0 = T()
       
       or(x,T()) = 2x >= 0 = T()
       
       imply(x,y) = 2x + y >= 2x + y = or(not(x),y)
      problem:
       or(x,T()) -> T()
       imply(x,y) -> or(not(x),y)
      Matrix Interpretation Processor: dim=1
       
       interpretation:
        [not](x0) = x0,
        
        [or](x0, x1) = 4x0 + 2x1 + 1,
        
        [imply](x0, x1) = 5x0 + 2x1 + 1,
        
        [T] = 2
       orientation:
        or(x,T()) = 4x + 5 >= 2 = T()
        
        imply(x,y) = 5x + 2y + 1 >= 4x + 2y + 1 = or(not(x),y)
       problem:
        imply(x,y) -> or(not(x),y)
       Matrix Interpretation Processor: dim=1
        
        interpretation:
         [not](x0) = x0,
         
         [or](x0, x1) = x0 + x1,
         
         [imply](x0, x1) = x0 + x1 + 1
        orientation:
         imply(x,y) = x + y + 1 >= x + y = or(not(x),y)
        problem:
         
        Qed


1:and(x,T()) -> x
  and(x,F()) -> F()
  and(x,y) -> and(y,x)
  and(and(x,y),z) -> and(x,and(y,z))
  AT confluence processor
   Complete TRS T' of input TRS:
   and(x,T()) -> x
   and(x,F()) -> F()
   and(T(),y) -> y
   and(F(),y) -> F()
   and(x,y) -> and(y,x)
   and(and(x,y),z) -> and(x,and(y,z))
   
    T' = (P union S) with
   
    TRS P:and(x,y) -> and(y,x)
          and(and(x,y),z) -> and(x,and(y,z))
   
    TRS S:and(x,T()) -> x
          and(x,F()) -> F()
          and(T(),y) -> y
          and(F(),y) -> F()
   
   S is left-linear and P is reversible.
   
    CP(S,S) = 
   T() = T(), F() = F()
   
    CP(S,P union P^-1) = 
   x = and(T(),x), and(x,y) = and(x,and(y,T())), and(x,z) = and(x,and(T(),z)), 
   y = and(T(),y), and(x,y) = and(and(x,y),T()), F() = and(F(),x), F() = 
   and(x,and(y,F())), and(F(),z) = and(x,and(F(),z)), F() = and(F(),y), 
   and(x,F()) = and(and(x,y),F()), y = and(y,T()), and(y,z) = and(T(),and(y,z)), 
   x = and(x,T()), and(y,z) = and(and(T(),y),z), and(x,z) = and(and(x,T()),z), 
   F() = and(y,F()), and(F(),z) = and(F(),and(y,z)), F() = and(x,F()), 
   F() = and(and(F(),y),z), and(x,F()) = and(and(x,F()),z)
   
   
    PCP_in(P union P^-1,S) = 
   
   
   We have to check termination of S:
   
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [and](x0, x1) = 5x0 + 5x1,
     
     [T] = 5,
     
     [F] = 0
    orientation:
     and(x,T()) = 5x + 25 >= x = x
     
     and(x,F()) = 5x >= 0 = F()
     
     and(T(),y) = 5y + 25 >= y = y
     
     and(F(),y) = 5y >= 0 = F()
    problem:
     and(x,F()) -> F()
     and(F(),y) -> F()
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [and](x0, x1) = 2x0 + 4x1,
      
      [F] = 2
     orientation:
      and(x,F()) = 2x + 8 >= 2 = F()
      
      and(F(),y) = 4y + 4 >= 2 = F()
     problem:
      
     Qed