Question

Using the blocks language, provide a Euler circle expressing the relations between logical equivalence, first-order equivalence, and tautological equivalence. Make sure to include at least one pair of sentences in each circle as an example

Answer 1

Answer:-

In any case, we're interested in semantic arguments for the truth of sentences. In the propositional case, we might ask for a semantic justification of the formula P→(P∨Q)P→(P∨Q). To demonstrate this with truth tables is straightforward, if somewhat tedious. To take the semantic approach, however, we look at the connectives in the sentence and what they mean.

(P∧R)→(P∨Q)(P∧R)→(P∨Q) is a conditional, and a conditional is true when the truth of the antecedent implies the truth of the consequent.

• The antecedent is P∧RP∧R, which is a conjunction. A conjunction is true if and only if both of its conjuncts are true. Its conjuncts are PP and QQ.
• The consequent is P∨QP∨Q, which is a disjunction. A disjunction is true if and only if at least one of its disjuncts are true. Its disjuncts are PP and QQ.

So, does the truth of the antecedent imply the truth of the consequent? The truth of the antecedent requires that PP be true, which means that at least one of PP and QQ is true, so the consequent must also be true. Therefore, the conditional is true.

A first order example

This type of reasoning carries over to the first-order case, but we must handle a few more types of sentences. Particularly, a first-order interpretation fixes a non-empty domain of discourse and maps each constant symbol to an element of the domain, each nn-ary relation symbol to a nn-ary relation on the domain. An interpretation II satisfies an expression P(a)P(a) if and only if the set that the interpretation maps PP to contains the individual that the interpretation maps aa to, i.e., if I(a)∈I(P)I(a)∈I(P). II satisfies a universal quantification ∀x.ϕ(x)∀x.ϕ(x) if for it satisfies ϕ(x)ϕ(x) for every xx in the domain, and ∃x.ϕ(x)∃x.ϕ(x) if it satisfies ϕ(x)ϕ(x) for some xx in the domain.

Let's consider the sentence ∀x.(P(x)∨¬P(x))∀x.(P(x)∨¬P(x)). This is a first-order validity. Why? Well, it is true if for each xx in the domain, P(x)∨¬P(x)P(x)∨¬P(x) is true. This is true if either P(x)P(x) is true or ¬P(x)¬P(x) is true.P(x)P(x) is true if xx is in I(P)I(P). ¬P(x)¬P(x) is true if P(x)P(x) is false. P(x)P(x) is false if x∉I(P)x∉I(P). Since xx must be in I(P)I(P) or not in I(P)I(P), at least one of these cases must be true, and so the sentence is true.

As another example, consider ∀x.P(x)→∃x.P(x)∀x.P(x)→∃x.P(x). This is true if the truth of ∀x.P(x)∀x.P(x) implies the truth of ∃x.P(x)∃x.P(x). Without going through all the low level details, ∀x.P(x)∀x.P(x) is true if every xx in the domain is in I(P)I(P). Since the domain is non-empty, that means that there is at least one xx such that x∈I(P)x∈I(P), and that is enough to make ∃x.P(x)∃x.P(x) true. Therefore, the truth of the antecedent implies the truth of the consequent, so the conditional is true.

Remarks

These kinds of explanations can be rather tedious in some ways, especially because they sometimes seem so obvious. We may be tempted to say, “well of course P∧QP∧Q being true implies that PP is true and QQ is true; that's what ∧∧ means!” But that's the whole point—logic systems have two sides: a syntactic side and a semantic side, and we have to be able to see the connection between the two. We tend to be pretty good at this when the syntax has symbols the meanings of which we understand, but with more complicated languages, it is important to be able to do this rigorously. Other times, we might be interested in exploring different possible semantics for languages. For instance, in standard first-order logic, an interpretation is based on a non-empty domain of discourse. If we consider a different semantics that allows empty domains of discourse, then semantic analysis will show that ∀x.(P(x)∨¬P(x))∀x.(P(x)∨¬P(x)) is a validity under both kinds of semantics, but ∃x.(P(x)∨¬P(x))∃x.(P(x)∨¬P(x)) is a validity only in the semantics that requires a non-empty domain of discourse. This means that sentences like ∀x.(P(x)∨¬P(x))→∃x.(P(x)∨¬P(x))∀x.(P(x)∨¬P(x))→∃x.(P(x)∨¬P(x)) will be true under one semantics and not the other. This means that the same set of inference rules will be sound for one semantics, but unsound for the other. We can only discover this if we can reason about these sentences semantically

To say that two propositions are “logically equivalent” is to say that they are true or false in exactly the same circumstances. We can show this by the use of truth tables. To say that two propositions are true in the same circumstances is just to say that they have the same truth-value (i.e., truth or falsity) for any given assignment of truth values to their atomic components. When we create a truth table, we are simply evaluating a statement (or a group of statements) for all such possible assignments. As an example, let’s start with the example from class. I said that “unless” could be translated as either “if not” or “or,” because (~Q ⊃ P) was logically equivalent to (P v Q). Let’s construct a truth table to prove this. First, we list on the top line of the table all of the atomic components (in this case, “P” and “Q”), and then each of the two statements we are evaluating for logical equivalence (in this case “(~Q ⊃ P)” and (P v Q)). Thus, P Q (~ Q ⊃ P) (P v Q) Then, we fill in the all the possible truth value assignment to the atomic components, as listed on the left (left of the double lines) side of the table. Thus, P Q (~ Q ⊃ P) (P v Q) 0 0 0 1 1 0 1 1 Note again that this list contains all the possible truth value assignments for a combination of two atomic propositions. We now move from the left side of the table to the right. For each “P” and “Q” on the right, we write the truth value assignment we find on the left. So, for the top line, we put a “0” under both “P” and “Q” on the right, while on the second line we put a “0” under each “P” and a “1” beneath each “1.” Thus: P Q (~ Q ⊃ P) (P v Q) 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 1 1 1 There are two propositions on the right part of the table, (~Q ⊃ P) and (P v Q). Before we can evaluate the truth value of (~Q ⊃ P), we must determine the value of ~Q for each line of the table. So, we do that next, entering this value for each line directly underneath the “~” sign. Thus: P Q (~ Q ⊃ P) (P v Q) 0 0 1 0 0 0 0 0 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 0 1 1 1 1 Now that we know the value of ~Q, we can determine the value of (~Q ⊃ P). Thus: P Q (~ Q ⊃ P) (P v Q) 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 1 1 Note that in this last move, we are applying the truth table for “⊃” to the values under “P” and under the “~” column (to the left of the “Q” column) because we want to know the value of “(~Q ⊃ P).” This gives us the truth value of (~Q ⊃ P) for all possible truth value assignments to all of its atomic components. Our last step is to calculate the truth value of (P v Q) by filling in the truth values underneath the “v” column. This gives us: P Q (~ Q ⊃ P) (P v Q) 0 0 1 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 Now we need only examine the results. We have considered the truth value of each statement for every possible combination of truth values to their atomic components. We find that for each assignment of truth values to these components (i.e., on each line), each compound proposition has the same truth value. (Look at the two columns in bold.) This means that the two propositions are true and false in exactly the same circumstances, i.e., that they are logically equivalent. Just for fun, let’s consider another logically equivalent pair of statements, “(P ⊃ Q)” and “(~Q ⊃ ~P).” P Q (P ⊃ Q) (~ Q ⊃ ~ P) 0 0 0 1 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 0 1 0 0 0 1 1 1 1 1 1 0 1 1