Show that the following proposition is a tautology [ ( x + r ) ^ ( Z + r ) ^ ( - x - 2 )
To show that a proposition is a tautology, we need to prove that it is always true, regardless of the truth values assigned to its variables. In this case, we have the proposition:
P = (x + r) ∧ (Z + r) ∧ (-x - 2)
To prove that this proposition is a tautology, we can use proof by contradiction. We assume that the proposition is not a tautology and show that it leads to a contradiction. So, let's assume that there is a truth assignment that makes P false.
Since P is a conjunction (using the symbol "∧" which represents logical "and"), for P to be false, any one of its conjuncts must be false. Let's examine each conjunct separately:
1. (x + r):
Let's consider a truth assignment that makes this conjunct false. Since addition is always commutative, we can rewrite this as (r + x). Assuming this conjunct is false, we have r + x = false. However, there is no way for the sum of two real numbers to be false. Therefore, this conjunct must always be true.
2. (Z + r):
Again, let's consider a truth assignment that makes this conjunct false. Rewriting it as (r + Z), we assume r + Z = false. Similarly, there is no way for the sum of a real number and an integer to be false. Thus, this conjunct is always true.
3. (-x - 2):
Finally, let's examine the negation of this conjunct. Assuming -x - 2 = false, we have -x = 2. This implies that x = -2, which is a specific solution, not a general contradiction. Therefore, this conjunct is sometimes true.
Based on our analysis, all the conjuncts of the proposition P are always true, except for the third one, which is sometimes true. This means that there exists a truth assignment that makes P false, so P cannot be a tautology.