1. More axioms
Disjunction
Copyright © 2007 Curt Hill

2. Disjunction has the following properties:
Associative
p  q  r  p  (q  r)  (p  q)  r
Symmetric
p  q  q  p
Idempotent
ppp
Distributes over equivalence
p  (q  r)  p  q  p  r
Higher precedence than 
Copyright © 2007 Curt Hill

3. First Disjunction Theorem
We can prove that it distributes over itself p  (q  r)  (p  q)  (p  r)
So lets do so
We may start with one side and argue to the other
We may start with entire expression and argue to a theorem or axiom
We will start with whole thing
Copyright © 2007 Curt Hill

4. Supposition
p  (q  r)  (p  q)  (p  r)
Copyright © 2007 Curt Hill

5. First step
p  (q  r)  (p  q)  (p  r) <Disjunction is associative> p  q  r  (p  q)  (p  r)
Copyright © 2007 Curt Hill

6. Second step
p  (q  r)  (p  q)  (p  r) <Disjunction is associative> p  q  r  (p  q)  (p  r) <Disjunction is idempotent> p  p  q  r  (p  q)  (p  r)
Copyright © 2007 Curt Hill

7. Third step
p  (q  r)  (p  q)  (p  r) <Disjunction is associative> p  q  r  (p  q)  (p  r) <Disjunction is idempotent> p  p  q  r  (p  q)  (p  r) <Disjunction is symmetric> p  q  p  r  (p  q)  (p  r)
Copyright © 2007 Curt Hill

8. Fourth step
p  (q  r)  (p  q)  (p  r) <Disjunction is associative> p  q  r  (p  q)  (p  r) <Disjunction is idempotent> p  p  q  r  (p  q)  (p  r) <Disjunction is symmetric> p  q  p  r  (p  q)  (p  r) <Disjunction is associative> (p  q)  (p  r)  (p  q)  (p  r)
Copyright © 2007 Curt Hill

9. Last step
p  (q  r)  (p  q)  (p  r) <Disjunction is associative> p  q  r  (p  q)  (p  r) <Disjunction is idempotent> p  p  q  r  (p  q)  (p  r) <Disjunction is symmetric> p  q  p  r  (p  q)  (p  r) <Disjunction is associative> (p  q)  (p  r)  (p  q)  (p  r) <x  x, where x is (p  q)  (p  r) > true
Copyright © 2007 Curt Hill

10. Aside on Proofs The above proof was much too elaborate
I made it detailed as an introduction
In most cases we will collapse several steps into one citing more than one property
Also the last step might be omitted because it is obvious
When in doubt err on the side of being detailed
Copyright © 2007 Curt Hill

11. Zero of disjunction is true
P  true  true
Ready? Lets try this one.
We will argue from left (more complicated) to right (less complicated) across the equivalence
Copyright © 2007 Curt Hill

12. Starting point
P  true
Copyright © 2007 Curt Hill

13. First Step P  true <Since qq is true, substitute> p  (qq)
Copyright © 2007 Curt Hill

14. Second Step
P  true <Since qq is true, substitute> p  (qq) <Distributives over equivalence> (p  q)  (p  q)
Copyright © 2007 Curt Hill

15. Conclusion
P  true <Since qq is true, substitute> p  (qq) <Distributives over equivalence> (p  q)  (p  q) <Reflexivity of equivalence> True
Copyright © 2007 Curt Hill

16. Aside So far we have started with several axioms on the disjunction:
Associative and Symmetric
Idempotent
ppp
Distributes over equivalence
p  (q  r)  p  q  p  r
Higher precedence than 
Proven several theorems:
Distributes over itself p  (q  r)  (p  q)  (p  r)
Zero: P  true  true
Copyright © 2007 Curt Hill

17. Yet We have said nothing about what it does
No truth table
No Venn diagram
The same may said for equivalence
Does this bother us? No
Copyright © 2007 Curt Hill

18. What else is missing? Plenty! The false is the unit of disjunction
However, we cannot yet state this because we have not yet defined false or negation
Instead we will slowly work through these operators and axioms
Copyright © 2007 Curt Hill