MHB Proving ((p->~q)∧q)->~p as a Tautology - Abdullah's FB Question

  • Thread starter Thread starter alyafey22
  • Start date Start date
AI Thread Summary
Abdullah's question focuses on proving that the expression ((p→¬q)∧q)→¬p is a tautology. The proof begins by applying the commutative law to rewrite the expression. It utilizes the equivalence p→q ≡ ¬p∨q to transform the expression into a more manageable form. By applying the distributive law and recognizing that q∧¬q is false, the expression simplifies to (q∧¬p)→¬p. Ultimately, the proof concludes that the expression is always true, confirming it as a tautology.
alyafey22
Gold Member
MHB
Messages
1,556
Reaction score
2
Abdullah asked the following question

prove that $$((p\to \neg q) \land q) \to \neg p$$

is a Tautology .
 
Mathematics news on Phys.org
By the commutative law we can rewrite as

$$(q \land (p\to \neg q) ) \to \neg p$$

First we need to know that

$$\tag{1}p\to q \equiv \, \neg p \lor q$$

Using this we get

$$q \land (\neg p \lor \neg q)\to \neg p $$

By distributive law

$$ (q \land \neg p) \lor (q \land \neg q)\to \neg p $$

Since

$$q \land \neg q \equiv F \,\,\, , \,\,\, q \lor F \equiv q$$

So we have

$$ (q \land \neg p)\to \neg p $$

Using (1) again

$$ \neg (q \land \neg p) \lor \neg p \equiv \neg q \lor p \lor \neg p\equiv T$$

Using De Morgan law .
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top