+3 votes
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Use Modus ponens $(A, A → B |= B)$ or resolution to show that the following set is inconsistent:

1. $Q(x) → P (x) \vee \sim R (a)$
2. $R (a) \wedge \sim Q(a)$
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.

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edited ago | 468 views

## 1 Answer

+1 vote

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$\because$ $x$ 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. $(R(a)\ \Lambda \sim Q(a))$
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

(as it will give True if $Q(a)$ is false and vice versa.

$\therefore$ always true/always false is not possible).

answered by Active (2k points)
edited ago by
0

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