All-versus-nothing proof of Bell’s
theorem for two observers
Part I: Proofs of Bell’s theorem
“without inequalities” (for qubits).
Beating EPR using their own weapons
(perfect correlations)
(Mermin’s version of)
Greenberger, Horne and Zelinger’s proof
GHZ 
1
  
2
X 1Y2Y3  1
Y1 X 2Y3  1
Y1Y2 X 3  1
X 1 X 2 X 3  1

Hardy’s proof
Any pure two-qubit entangled, but not maximally
entangled, state (Hardy state) can be written as
Logical argument:
  c  A  B   c   A  B   c  A  B   c   A  B 
 d   A  b   d  A  b   d  A  b 
 f   a  B   f   a  B   f  a  B 
 g   a  b   g  a  b   g  a  b 
Therefore,
P  A  1, B  1  c 
2
P b  1 A  1  1
P a  1 B  1  1
P a  1, b  1  0
“A veces” = “Sometimes”
“Siempre” = “Always”
“Nunca” = “Never”
FAQ
What if the measurement of x1x2 changes v(z1z2)?
Do you need to assume that v(z1z2) does not change when Alice measures the
compatible observable x1x2?
Do you need to assume noncontextuality in addition to EPR’s elements of reality?
No. The assumption of EPR's elements of reality prevents that v(z1z2) changes
when x1x2 is measured on Alice's subsystem (qubits 1 and 2).
The key is to note that v(z1z2) can be predicted with certainty from space-like
separated measurements on Bob's subsystem (qubits 3 and 4).
Bob's prediction for v(z1z2) is always confirmed: not only when z1z2 is measured
alone, but also when z1z2 and x1x2 are measured together, and when z1z2 is
measured after x1x2.
The beauty contest
The beauty contest
The beauty contest
Part II: Experimental all-versusnothing proof of Bell’s theorem for two
observers