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

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The discussion revolves around two proofs demonstrating the logical equivalence of p and its negation, p<=>~p, leading to contradictory outcomes. The first proof concludes with a contradiction (~p^p), while the second results in an identity (p=>p). The proofs illustrate that the assumption p<=>~p is a self-contradiction, as it can derive both a contradiction and a tautology. Participants debate the implications of syntactically proving that any formula logically implies Modus Ponens and the nature of logical implications between formulas. The conversation emphasizes the relationship between contradictions and tautologies in logical derivations.
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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
 
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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 $\bot$.
 
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.
 
Evgeny.Makarov said:
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
 
Ackbach said:
Or, put another way, your assumption is a self-contradiction. .

I was trying to prove that syntactically
 
solakis said:
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.

solakis said:
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 $B$).
 

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