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So I'm proving a tautology with logical equivalents, but as I get to the end I'm not getting a truth. The question states 'Prove that each statement is a tautology' which leads me to believe that the statement is in fact true. Was wondering if anyone could take a look at my work and tell me if they spot any errors? I've tried a couple different ways, but this is the closest I've been able to get.

STATEMENT: [(p v q) ^ (p -> r) ^ (q -> r)] -> r

L.E.1 : [(p v q) ^ ('p v q) ^ ('q v r)] -> r ////implication

L.E.2: '[(p v q ) ^ ('p v q) ^ ('q v r)] v r /////implication

L.E.3: ['(p v q) v '('p v q) v '('q v r)] v r ///// demorgans

L.E.4: [('p ^ q) v (p ^ q) v (q ^ r)] v r/////demorgans

L.E.5: [('p ^ q) v (q ^ p) v (q ^ r)] v r//////communicative

L.E.6: [('p ^ q) v q ^ (p v r)] v r///////////distributive

L.E.7: [(q ^ 'p) v (q ^ (p v r)] v r///////////communicative

L.E.8: [(q ^ ('p v (p v r))] v r/////////////distributive

L.E.9: [(q ^ (('p v p) v r)] v r//////////////distributive

L.E.10: [q ^ (T v r) ] v r //////////////////complement

L.E.11: [q ^ T] v r ///////////////////////identity

L.E.12: q v r //final form.

As you can see, I am unable to get it to evaluate to true, but merely 'q v r'

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    Your second application of De Morgan's law is wrong. – 2017-01-19
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    L.E.1 has a typo: a $q$ instead of an $r$. – 2017-01-19
  • 1
    You can use ` ` or `$\quad$` to create spaces. Also, please learn LaTeX to post on this site. Google StackExchange MathJax guide if you didn't see it in the instructions when you posted your question. – 2017-01-20
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    I was wondering how to format. Thank you for this information – 2017-01-22

1 Answers 1

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STATEMENT: [(p v q) ^ (p -> r) ^ (q -> r)] -> r

L.E.1 : [(p v q) ^ ('p v $\color{red}{q}$) ^ ('q v r)] -> r $\color{red}{\ Typo: should\ be\ r\ not\ q}$

L.E.2: '[(p v q ) ^ ('p v q) ^ ('q v r)] v r

L.E.3: ['(p v q) v '('p v q) v '('q v r)] v r

L.E.4: [('p ^ $\color{red}{q}$) v (p ^ $\color{red}{q}$) v (q ^ $\color{red}{r}$)] v r$\color{red}{\ Forgot\ to\ negate\ second\ variable\ in\ the\ brackets\ when\ using\ demorgans}$

. . .

L.E.12: q v r //final form.

Apart from those mistakes the rest is ok (except of the course the final answer).

This is what I obtained when expanding your statement:

[(p + q) ^ (p -> r) ^ (q -> r)] -> r

$\bar p\bar q$+p$\bar r$+q$\bar r$+r

$\bar p\bar q$+$\bar r$[p+q] +r

Let X=[p+q]

$\bar p\bar q$+$\bar r$X +r

$\bar p\bar q$+$\bar r$X +rX+r$\bar X$

$\bar p\bar q$+X(1)+r$\bar X$

But as X=[p+q] $\bar X=\bar p\bar q$ so

$\bar p\bar q$+[p+q]+r$\bar p\bar q$

$\bar p\bar q$(1+r)+[p+q]

p+q+$\bar p\bar q$

Which is a Tautology

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    I figured it out, but this is correct nonetheless so I'll give you the credit. – 2017-01-22