Quantum Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  QLE Home  >  Th. List  >  dp15lemb GIF version

Theorem dp15lemb 1153
 Description: Part of proof (1)=>(5) in Day/Pickering 1982.
Hypotheses
Ref Expression
dp15lema.1 d = (a2 ∪ (a0 ∩ (a1b1)))
dp15lema.2 p0 = ((a1b1) ∩ (a2b2))
dp15lema.3 e = (b0 ∩ (a0p0))
Assertion
Ref Expression
dp15lemb ((a0a1) ∩ (eb1)) ≤ (((a0d) ∩ (eb2)) ∪ ((a1d) ∩ (b1b2)))

Proof of Theorem dp15lemb
StepHypRef Expression
1 dp15lema.1 . . 3 d = (a2 ∪ (a0 ∩ (a1b1)))
2 dp15lema.2 . . 3 p0 = ((a1b1) ∩ (a2b2))
3 dp15lema.3 . . 3 e = (b0 ∩ (a0p0))
41, 2, 3dp15lema 1152 . 2 ((a0e) ∩ (a1b1)) ≤ (db2)
54ax-arg 1151 1 ((a0a1) ∩ (eb1)) ≤ (((a0d) ∩ (eb2)) ∪ ((a1d) ∩ (b1b2)))
 Colors of variables: term Syntax hints:   = wb 1   ≤ wle 2   ∪ wo 6   ∩ wa 7 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-ml 1120  ax-arg 1151 This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131 This theorem is referenced by:  dp15lemc  1154
 Copyright terms: Public domain W3C validator