Proofs and Inferences in Proving Propositional Theorem

Last Updated : 28 Feb, 2022

This article discusses how to use inference rules to create proof—a series of conclusions that leads to the desired result. The most well-known rule is known as Modus Ponens (Latin for affirming mode) and is expressed as

$\frac{\alpha \Rightarrow \beta, \quad \alpha}{\beta}$

Inferences in Proving Propositional Theorem

The notation signifies that the sentence may be deduced whenever any sentences of the type are supplied. If $\text { ( WumpusAhead } \wedge \text { WumpusAlive) } \Rightarrow \text { Shoot }$ and $\text { (WumpusAhead } \wedge \text { WumpusAlive) }$ are both supplied, Shoot may be deduced.

And-Elimination is another helpful inference rule, which states that any of the conjuncts can be inferred from conjunction:

$\frac{\alpha \wedge \beta}{\alpha}$

WumpusAlive can be deduced from $\text { (WumpusAhead } \wedge \text { WumpusAlive) }$ , for example. One may readily demonstrate that Modus Ponens and And-Elimination are sound once and for all by evaluating the potential truth values of $\alpha$ and $\beta$ . These principles may then be applied to each situation in which they apply, resulting in good conclusions without the necessity of enumerating models.

$\begin{aligned} (\alpha \wedge \beta) & \equiv(\beta \wedge \alpha) \quad \text { commutativity of } \wedge \\ (\alpha \vee \beta) & \equiv(\beta \vee \alpha) \quad \text { commutativity of } \vee \\ ((\alpha \wedge \beta) \wedge \gamma) & \equiv(\alpha \wedge(\beta \wedge \gamma)) \quad \text { associativity of } \wedge \\ ((\alpha \vee \beta) \vee \gamma) & \equiv(\alpha \vee(\beta \vee \gamma)) \quad \text { associativity of } \vee \\ \neg(\neg \alpha) & \equiv \alpha \quad \text { double-negation elimination } \\ (\alpha \Rightarrow \beta) & \equiv(\neg \beta \Rightarrow \neg \alpha) \quad \text { contraposition } \\ (\alpha \Rightarrow \beta) & \equiv(\neg \alpha \vee \beta) \quad \text { implication elimination } \\ (\alpha \Leftrightarrow \beta) & \equiv((\alpha \Rightarrow \beta) \wedge(\beta \Rightarrow \alpha)) \quad \text { biconditional elimination } \\ \neg(\alpha \wedge \beta) & \equiv(\neg \alpha \vee \neg \beta) \quad \text { De Morgan } \\ \neg(\alpha \vee \beta) & \equiv(\neg \alpha \wedge \neg \beta) \quad \text { De Morgan } \\ (\alpha \wedge(\beta \vee \gamma)) & \equiv((\alpha \wedge \beta) \vee(\alpha \wedge \gamma)) \quad \text { distributivity of } \wedge \text { over } \vee \\ (\alpha \vee(\beta \wedge \gamma)) & \equiv((\alpha \vee \beta) \wedge(\alpha \vee \gamma)) \quad \text { distributivity of } \vee \text { over } \wedge \end{aligned}$

The equations above show all of the logical equivalences that can be utilized as inference rules. The equivalence for biconditional elimination, for example, produces the two inference rules.

$\frac{\alpha \Leftrightarrow \beta}{(\alpha \Rightarrow \beta) \wedge(\beta \Rightarrow \alpha)} \quad \text { and } \quad \frac{(\alpha \Rightarrow \beta) \wedge(\beta \Rightarrow \alpha)}{\alpha \Leftrightarrow \beta}$

Some inference rules do not function in both directions in the same way. We can’t, for example, run Modus Ponens in the reverse direction to get $\alpha \Rightarrow \beta$ and $\alpha \text{ from } \beta$ .

Let’s look at how these equivalences and inference rules may be applied in the wumpus environment. We begin with the knowledge base including R1 through R5 and demonstrate how to establish $\neg P_{1,2}$ i.e. that [1,2] does not include any pits. To generate R6, we first apply biconditional elimination to R2:

$R_{6}: \quad\left(B_{1,1} \Rightarrow\left(P_{1,2} \vee P_{2,1}\right)\right) \wedge\left(\left(P_{1,2} \vee P_{2,1}\right) \Rightarrow B_{1,1}\right)$

After that, we apply And-Elimination on R6 to get $R_{7}: \quad\left(\left(P_{1,2} \vee P_{2,1}\right) \Rightarrow B_{1,1}\right)$

For contrapositives, logical equivalence yields $R_{8}: \quad\left(\neg B_{1,1} \Rightarrow \neg\left(P_{1,2} \vee P_{2,1}\right)\right)$

With R8 and the percept $R_{4} \text { (i.e., } \neg B_{1,1} \text { ) }$ , we can now apply Modus Ponens to get $R_{9}: \quad \neg\left(P_{1,2} \vee P_{2,1}\right)$ .

Finally, we use De Morgan’s rule to arrive at the following conclusion: $R_{10}: \quad \neg P_{1,2} \wedge \neg P_{2,1}$

That is to say, neither [1,2] nor [2,1] have a pit in them.

We found this proof by hand, but any of the search techniques may be used to produce a proof-like sequence of steps. All we have to do now is define a proof problem:

Initial State: the starting point for knowledge.
Actions: the set of actions is made up of all the inference rules that have been applied to all the sentences that fit the inference rule’s upper half.
Consequences: Adding the statement to the bottom part of the inference rule is the result of an action.
Objective: The objective is to arrive at a state that contains the phrase we are attempting to verify.

As a result, looking for proofs is a viable alternative to counting models.

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