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I am studying for an exam to be taken next Friday and I am only stuck on one question in my revision guide, which I simply do not understand. (I even have the answers as well)

The question:

Show, by producing a counter-model in each case, that neither of the following is a valid argument:

(a) $$\forall x(G(x) \to H(x)),G(a) \mid= \forall xH(x)$$

(b) $$H(a) \to \forall xG(x), \lnot(G(a) \lor G(b)) \mid= \lnot\exists xH(x)$$

And the answer:

We need to find a counter-model, i.e. a model in which the formulas on the left-hand side of $\mid=$ are true and that on the right-hand side is false.

(a) Take $M = (\{1,2\},I)$, where $I(a) = 1, I(G) = \{1\}$ and $I(H) = \{1\}$. Then $\forall x(G(x) \to H(x))$ holds in $M$, $G(a)$ does too, but $H(2)$ does not hold. So $\forall xH(x)$ does not hold in $M$.

(b) Take $M = (\{1,2\},I)$, where $I(a) = 1$, $I(b) = 2$, $I(G) = 0$ and $I(H) = {2}$. Then $H(a) \to \forall xG(x)$ is satisfied and so is $\lnot(G(a)\lor G(b))$. But $H(b)$ holds, so it is not true that $\lnot \exists xH(x)$ holds.

What I know:

When answering these questions before looking at the answers I used the Predicate Tableux method to evaluate that a false outcome was possible when all statements were true.

I really have no idea how these answers were concluded. Is there a way to produce such a answer derived from a completed tableaux showing that a false outcome is possible when all statements are true? (The reason I ask is because I find tableaux rather easy and would like to stick to it to not cause ambiguity in the exam)

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As stated in your problem, in order to prove that an argument is not valid, we have to produce a counter-model.

To use the tableaux method for finding a counter-model to :

$∀x(G(x) → H(x)), G(a) \vDash ∀xH(x)$ --- (*)

we have to consider the formula :

$∀x(G(x) → H(x)) \land G(a) \land \lnot ∀xH(x)$ --- (§)

and "run" the tableaux. If it will not close, the "open" path will give us a counter-example for (*).

This because the open path will define a "model" satisfying the formula (§).

But, in general, we have that $p \to q$ is equivalent to $\lnot (p \land \lnot q)$.

Thus, to find a model for $(p \land \lnot q)$ means that it is satisfiable; but if it is satisfiable, then its negation, i.e. $p \to q$ is not valid.

Finally, we have that $\vDash p \to q$ iff $p \vDash q$.

Thus, to prove that (§) is satisfiable is equivalent to prove that (*) is not a valid argument, i.e. that the conclusion : $∀xH(x)$, does not follow from the premises : $∀x(G(x) → H(x)), G(a)$.


Example

1) $∀x(G(x) → H(x)) \land G(a) \land \lnot ∀xH(x)$

2) $∀x(G(x) → H(x))$ --- from 1) by rule for $\land$

3) $G(a)$ --- as 2)

4) $\lnot ∀xH(x)$ --- as 2)

5) $\lnot H(b)$ --- from 4) by rule for $\lnot \forall$ : $b$ new

6) $G(a) → H(a)$ --- from 2) by rule for $\forall$ with $a$

7) $G(b) → H(b)$ --- from 2) by rule for $\forall$ with $b$

Now we use the $\to$ rule with 6), generating two branches : $8_L$ and $8_R$ :

$8_L$) $\lnot G(a)$ --- this path closes with 3) : $\times$

$8_R$) $H(a)$

Now we use again the $\to$ rule with 7), generating two branches : $9_L$ and $9_R$ :

$9_L$) $\lnot G(b)$ --- OPEN

$9_R$) $H(b)$ --- this path closes with 5) : $\times$.

Conclusion : the path through 1)-7), $8_R$), $9_L$) is open and thus all the formulae on it (including 1)) are satisfiable

The satisfying model must be manufactured starting from atoms :

$G(a), \lnot H(b), H(a), \lnot G(b)$.

From this, the answer :

Take $\mathcal M = ( \{ 1,2 \} , \mathcal I)$, where $\mathcal I(a)=1, \mathcal I(b)=2, \mathcal I(G)= \{ 1 \}$ and $\mathcal I(H)= \{ 1 \}$.

We need a domain with two objects for the two parameters $a,b$ respectively (i.e. : $1,2$), such that the two subsets of the domain of $\mathcal M$ used to interpret the two predicates $G,H$ (i.e. $\mathcal I(G), \mathcal I(H)$) both contain the object assigned to $a$ (due to : $G(a)$ and $H(a)$); thus, we need :

$1 \in \mathcal I(G)$ and $1 \in \mathcal I(H)$.

We need also that the interpretation of $G$ does no contain the object assigned to $b$ (due to $\lnot G(b)$), i.e. :

$2 \notin \mathcal I(G)$.

Finally, we need that the interpretation of $H$ does no contain the object assigned to $b$ (due to $\lnot H(b))$, i.e. :

$2 \notin \mathcal I(H)$.

Now we can easily check that $\mathcal M \vDash \forall x(G(x) \to H(x))$ and that $\mathcal M \nvDash \forall xH(x)$.

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  • $\begingroup$ Thank you ever so much for this! Simply brilliant $\endgroup$ – Prolog Feb 22 '15 at 11:45

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