1,440 views

Use Modus ponens $(A, A → B |= B)$ or resolution to show that the following set is inconsistent:

1. $Q(x) \rightarrow P (x) \vee \sim R (a)$
2. $R (a) \vee \sim Q(a)$
3. $Q(a)$
4. $\sim P (y)$

where $x$ and $y$ are universally quantified variables, $a$ is a constant and $P, Q, R$ are monadic predicates.

1 comment

The second proposition is given as R(a)∧∼Q(a) instead of R(a)∨∼Q(a) in GO PDF 2020. Please correct it.

$\because$ $x$ and $y$ are universally quantified variable, we can write the given arguments as follows :-

1. $\forall x(Q(x)\rightarrow (P(x)\ V \sim R(a)))$
2. $(R(a)\ \vee \sim Q(a))$
3. $Q(a)$
4. $\forall y (\sim P(y))$

Now using Universal instantiation, $1.$ becomes

1. $Q(a)\rightarrow (P(a)\ V \sim R(a))$ where $a$ is an arbitrary constant given in question.

Similarly 4. becomes

1. $\sim P(a)$

Using Modus Ponens 5. and 3.

$Q(a)\rightarrow (P(a)\ V \sim R(a))$

$\underline {Q(a)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

1. $\therefore\ P(a)\ V \sim R(a)$

Using resolution 7. and 2.

$R(a)\ V \sim Q(a))$

$\underline { \sim R(a)\ V P(a))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

1. $\therefore\ P(a)\ V \sim Q(a)$

Using 6. and 8.

$P(a)\ V \sim Q(a)$

$\underline{\sim P(a) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$\therefore\ \sim Q(a)$

After applying appropriate rules of inference, at last we get $\sim Q(a)$, which is inconsistent with $(3)$ which requires $Q(a).$

why can't we further resolve Q(a) (obtained by resolution) with Q(a) (given in question)?

In the resolution step, isn't $2$ actually $R(a)$ AND NOT $Q(a)$? But you've written it as OR. I didn't get how that came about.
Yes I too don't find

Using resolution 7. and 2 step correct as in question it is R(a)R(a) AND NOT Q(a).

@ that won't change the result of the reduction