# Tag Info

4

You have the premise $(p\lor q)\lor r)$ which is a disjunction of disjunctions. You seek to conclude $p\lor(q\lor r)$ which is also a disjunction of disjunctions. Therefore your proof shall only use the rules of disjunction elimination and disjunction introduction. $$\def\fitch#1#2{~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{(p\lor q)\lor r}{\fitch{~}{\... 3 don't understand lines 9 and 10. What is that supposed to mean? Line 9 is by 'negation elimination'. From ~q~ and ~\lnot q~ you may infer ~\bot~. The symbol is the contradiction constant, also known as 'falsum' or 'bottom'. (From any statement and its own negation you may infer that you have a contradiction). Some systems call this 'contradiction ... 3 From$$\tag2 \operatorname{Loves}(b,a)$$introduce \exists to find$$\tag3\exists y\operatorname{Loves}(b,y)$$Specialize$$\tag1\forall x(\exists y\operatorname{Loves}(x,y)\to \forall z\operatorname{Loves}(z,x)$$to$$\tag4\exists y\operatorname{Loves}(b,y)\to \forall z\operatorname{Loves}(z,b)$$Apply modus ponens to (4) and (3) to find$$\forall z\...

2

I think disjunction elimination is your friend here. Namely, as a natural deduction, if you have as a starting assumption $(p \vee q) \vee r$, then you can have two subproofs, one of which that assume $p \vee q$, the other of which assumes $r$. If you can prove in each of those subproofs, under those assumptions, the desired consequence $p \vee (q \vee r)$, ...

2

I'm confused because from what I understand the ⊢ means that whenever the left-hand side is true, then the right hand side also is. That is the definition for semantic entailment, or models, $\models$. Syntactic entailment, or derives, $\vdash$, means that the the right-hand-side can be infered from the left-hand-side, using the syntactic proof system you ...

1

How exactly your proof needs to be will depend on the exact rules you will be using, in particular the existential quantifier might be tricky. A proof outline using the rules in the proof system at open logic project is as follows: Since your desired conclusion is a conditional, start by assuming the hypothesis $\exists x Qx$. Introduce a subproof with ...

1

The first proof uses the Principle of Explosion, which states that from a contradiction, anything follows. An alternative way to write this part of the proof would be like this: $(8)$ Assume $q$ $(9)$ Assume $\lnot r$ $(10)$ Contradiction between $(7)$ and $(8)$ $(11)$ Close assumption at $(9)$: contradiction implies $r$ (proof by contradiction) $(12)$ ...

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For your first question, I think they are using the Principle of Explosion, which says that from a contradiction, anything is provable. An argument for this might go as follows: assume that $p\wedge \neg p$ holds for some proposition $p$ and take any proposition $q$. Then, $p\rightarrow q$ is a true statement because $\neg p$ is true (by assumption). Then, ...

1

In your attempt, you used ($\lor$ Elim) to show $a\to b$, in order to conclude $\lnot b$, which is correct. However, it's unnecessary, we can just assume $a$, then reit $b$, so we have $a\to b$ hold. Then you assumed $\lnot a$, but stucked on showing that $a\to b$, a hint for this is that once we assume $a$ under $\lnot a$ it's a contradiction, use this and (...

1

To reach $A\&\lnot B$, prove $A$ and $\lnot B$ separately, then use conjunction introduction. To prove them use a reduction to absurdity and an indirect proof, respectively. Both require deriving a contradiction from an assumption, and the only other thing to contradict is the premise, which is a negation of a conditional), so derive that conditional......

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Others, particularly Graham Kemp, have given you excellent answers to your specific questions. But standing back a bit from the details, I suspect that you probably need to be doing more background homework about this kind of natural deduction system -- in particular, one deploying the absurdity constant. (As a general principle, if one textbook is leaving ...

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Answering your first question about the right hand side, assuming $p$ is true, we are left to prove $q \implies r$. If $q$ was false, you can't conclude anything about $r$. It can be either true or false! That's why we assume $p$ is true in the first place: if the whole relation is indeed true, assuming $p$ is true should lead us to conclude that \$q \implies ...

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