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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.

asked in Mathematical Logic by Veteran (52k points)
edited ago by | 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

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

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