Rules of Inference Help

**tokio** · March 7th 2009, 07:44 PM

Here is a riddle from A from ancient china, discovered in the recently revealed writings of Laong Dong.

If Ang lives the emperor, then Bao loves the emperor.
If Chang loves the emperor, then both Ang and Di love the emperor.
If both Bao and Di love the emperor, then both Ang and Eng love the emperor.

Eng does not love the emperor.
Therefore, Chang does not love the emperor.

Prove that the logic of this riddle is valid.

---------------------------

Here is what i have so far.

1. $a -> b$ Premise
2. $c->(a\wedge d)$ Premise
3. $(b \wedge d)->(a \wedge e)$ Premise
4. $-e$ Premise

-----------------
c Conclusion

Im stuck on how to get to the conclusion. I used Demorgan's rule for the premises to see if anything cancel out but cant seem to have any luck. Any help would be super, thanks.

**horan** · March 7th 2009, 08:29 PM

Here you go!

a := ang lives with the emperor
b := bao loves the emperor
c := chang loves the emperor
d := di loves the emperor
e := eng loves the emperor

$a \implies b \equiv \sim a \vee b \equiv T$
$c \implies (a \wedge d) \equiv \sim c \vee (a \wedge d) \equiv (a \vee \sim c) \wedge (d \vee \sim c) \equiv T$
$b \wedge d \implies a \wedge e \equiv \sim (b \wedge d) \vee (a \wedge e) \equiv T$

Want to show that $c \equiv F$

We know that e = F. Then $a \wedge e = F$ . But $\sim (b \wedge d) \vee (a \wedge e) \equiv T$ , so
we know that $\sim (b \wedge d) \equiv \sim b \vee \sim d \equiv T$ . Therefore, either b=F or d=F.

Next, we know that either ~c or a^d is true. We also know that ~a or b is true.

Case 1. Suppose b=F. Then ~a has to be true, and so a is false. Then (a^d) is false, so ~c is true, and so c is false.

Case 2. Suppose d=F. Then ~c has to be true, and so c is false.

Since c is false in both cases, Chang does not love the emperor.

**tokio** · March 7th 2009, 08:45 PM

Originally Posted by horan

Here you go!

a := ang lives with the emperor
b := bao loves the emperor
c := chang loves the emperor
d := di loves the emperor
e := eng loves the emperor

$a \implies b \equiv \sim a \vee b \equiv T$
$c \implies (a \wedge d) \equiv \sim c \vee (a \wedge d) \equiv (a \vee \sim c) \wedge (d \vee \sim c) \equiv T$
$b \wedge d \implies a \wedge e \equiv \sim (b \wedge d) \vee (a \wedge e) \equiv T$

Want to show that $c \equiv F$

We know that e = F. Then $a \wedge e = F$ . But $\sim (b \wedge d) \vee (a \wedge e) \equiv T$ , so
we know that $\sim (b \wedge d) \equiv \sim b \vee \sim d \equiv T$ . Therefore, either b=F or d=F.

Next, we know that either ~c or a^d is true. We also know that ~a or b is true.

Case 1. Suppose b=F. Then ~a has to be true, and so a is false. Then (a^d) is false, so ~c is true, and so c is false.

Case 2. Suppose d=F. Then ~c has to be true, and so c is false.

Since c is false in both cases, Chang does not love the emperor.

Hi, thanks, this is a litle different from how im taught to do this. We use rules like modus poens, tollens, hypthetical syllogism, etc. However, you ddi gave me an idea. Is this step logiclaly correct?

Im wondering what its called? Thanks.

Tokio

**horan** · March 7th 2009, 09:53 PM

Yes, the distributive law of logic allows you to do that, kind of how like the distributive law of multiplication allows you to do $3 * (2+5) = (3*2)+(3*5)$ .

**horan** · March 7th 2009, 09:55 PM

oh and the only reason you can set that equal to T is because it's a given statement we know is true.

**tokio** · March 8th 2009, 07:25 AM

Ah i see, you used Demorgan's rule first and then the distributive law, ok. After i used the distributive law i broke the two components up via simplification. So i had.

$(c V a) \wedge (-c V d)$

From there i used the rule of simplification and got

(-c V a)
(-c V d)

and then used resolution with the premise.

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