Proving p<=>~p: Two Contradictory and Identical Proofs Explained

[solakis1]

CANNOT FIGURE THIS OUT:

proof No1

1. p<=>~p.........assumption

2. (p<=>~p)<=>[(p=>~p)^(~p=>p)]......definition of (1)

3. (p<=>~p)=>[(p=>~p)^(~p=>p)].....2,and Biconditional Elimination

4. (p=>~p)^(~p=>p).........1,3 M.Ponens

5. (~p=>p)............4, Simplification

6. (p=>~p)............4,and using Simplification

7. ~~pvp............5, material implication

8. ~pv~p............6, material implication

9. pvp.............7. negation elimination

10. ~p.............8, idempontent law

11. p.............9, idempontent law

12. ~p^p............10,11 and Conjunctionproof No 2

1. p<=>~p.........assumption

2. (p<=>~p)<=>[(p=>~p)^(~p=>p)]......definition of (1)

3. (p<=>~p)=>[(p=>~p)^(~p=>p)].....2, Biconditional Elimination

4. (p=>~p)^(~p=>p).........1,3 M.Ponens

5. p=>p............4, hypothetical Syllogism

6. ~pvp............5, material implication

In the 1st proof we ended up with a contradiction ,while in the 2nd proof with an identity

 

[Evgeny.Makarov]

Let's define a formula $A$ to be stronger than a formula $B$ (or $B$ weaker than $A$) if $A\to B$ is a tautology, or, equivalently, $A\vdash B$. Then $\bot$ (the contradiction) is the strongest formula because $\bot\to B$ is true for any $B$. Conversely, $\top$ (the tautology) is the weakest formula because $A\to\top$ is true for any $A$. Derivation 2 shows that $p\leftrightarrow\neg p\vdash\top$, but this is obvious since every formula is stronger than $\top$. Derivation 1 shows that $p\leftrightarrow\neg p\vdash\bot$, which shows that $p\leftrightarrow\neg p$ is at least as strong as $\bot$ and, therefore, is equivalent to $\bot$.

 

[Ackbach]

Or, put another way, your assumption is a self-contradiction. Since you can prove anything whatsoever from a contradiction, it's not surprising that you got two such widely different results from a contradiction.

 

[solakis1]

because $A\to\top$ is true for any $A$

So how are you going to prove that any formula A logically implies M.PONENS for e.g

I mean syntactically not semantically ,like you did.

And then generally that any formula A logically implies any identity B

 

[solakis1]

Or, put another way, your assumption is a self-contradiction. .

I was trying to prove that syntactically

 

[Evgeny.Makarov]

So how are you going to prove that any formula A logically implies M.PONENS for e.g

I am not sure what you mean by "formula A logically implies M.PONENS". Modus Ponens is an inference rule, but only formulas can be logically implied.

And then generally that any formula A logically implies any identity B

I am not sure what you mean by "identity". If you mean tautology, then it is logically implied by definition. We say that $A$ logically implies $B$, written as $A\models B$, if $B$ is true in every interpretation in which $A$ is true. If you mean that $A$ derives $B$, written $A\vdash B$, for any tautology $B$ and any formula $A$, then this is the content of the completeness theorem (which shows $\vdash B$ for any tautology $B$).