YES

We show the termination of the TRS R:

  xor(x,F()) -> x
  xor(x,neg(x)) -> F()
  and(x,T()) -> x
  and(x,F()) -> F()
  and(x,x) -> x
  and(xor(x,y),z) -> xor(and(x,z),and(y,z))
  xor(x,x) -> F()
  impl(x,y) -> xor(and(x,y),xor(x,T()))
  or(x,y) -> xor(and(x,y),xor(x,y))
  equiv(x,y) -> xor(x,xor(y,T()))
  neg(x) -> xor(x,T())

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: and#(xor(x,y),z) -> xor#(and(x,z),and(y,z))
p2: and#(xor(x,y),z) -> and#(x,z)
p3: and#(xor(x,y),z) -> and#(y,z)
p4: impl#(x,y) -> xor#(and(x,y),xor(x,T()))
p5: impl#(x,y) -> and#(x,y)
p6: impl#(x,y) -> xor#(x,T())
p7: or#(x,y) -> xor#(and(x,y),xor(x,y))
p8: or#(x,y) -> and#(x,y)
p9: or#(x,y) -> xor#(x,y)
p10: equiv#(x,y) -> xor#(x,xor(y,T()))
p11: equiv#(x,y) -> xor#(y,T())
p12: neg#(x) -> xor#(x,T())

and R consists of:

r1: xor(x,F()) -> x
r2: xor(x,neg(x)) -> F()
r3: and(x,T()) -> x
r4: and(x,F()) -> F()
r5: and(x,x) -> x
r6: and(xor(x,y),z) -> xor(and(x,z),and(y,z))
r7: xor(x,x) -> F()
r8: impl(x,y) -> xor(and(x,y),xor(x,T()))
r9: or(x,y) -> xor(and(x,y),xor(x,y))
r10: equiv(x,y) -> xor(x,xor(y,T()))
r11: neg(x) -> xor(x,T())

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: and#(xor(x,y),z) -> and#(x,z)
p2: and#(xor(x,y),z) -> and#(y,z)

and R consists of:

r1: xor(x,F()) -> x
r2: xor(x,neg(x)) -> F()
r3: and(x,T()) -> x
r4: and(x,F()) -> F()
r5: and(x,x) -> x
r6: and(xor(x,y),z) -> xor(and(x,z),and(y,z))
r7: xor(x,x) -> F()
r8: impl(x,y) -> xor(and(x,y),xor(x,T()))
r9: or(x,y) -> xor(and(x,y),xor(x,y))
r10: equiv(x,y) -> xor(x,xor(y,T()))
r11: neg(x) -> xor(x,T())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = ((1,1),(1,0)) x1 + x2
        xor_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        and#_A(x1,x2) = ((0,0),(1,1)) x1 + x2
        xor_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1, p2

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