Return to main page
Additional Material
Truth set of a conjunction of two open statements
This is best explained with a specific example.
Every number in N9 receives two truth values from the open statements
x is in {2,3,4,5,6}, x is in {4,5,6,7,8}
They assign the following truth values to each number in N9:
1 FF,
2 TF,
3 TF,
4 TT,
5 TT,
6 TT,
7 FT,
8 FT,
9 FF
According to the discussion of conjunction in Chapter 1, a conjunction of two losgical statements is true if and only if both are true. This can restated in an abbreviated form of a table:
T and T → T
T and F → F
F and T → F
F and F → F
Therefore replace the above pairs of truth values with single truth values to get the table:
1 F,
2 F,
3 F,
4 T,
5 T,
6 T,
7 F,
8 F,
9 F
Consider now x is in {2,3,4,5,6} and x is in {4,5,6,7,8} to be a single open statement. This table shows the truth set of the conjunction to be {4,5,6}.
Truth set of an alternation of two open statements
This is best explained with a specific example.
Every number in N9 receives two truth values from the open statements
x is in {2,3,4,5,6}, x is in {4,5,6,7,8}
They assign the following truth values to each number in N9:
1 FF,
2 TF,
3 TF,
4 TT,
5 TT,
6 TT,
7 FT,
8 FT,
9 FF
According to the discussion of conjunction in Chapter 1, a conjunction of two losgical statements is true if and only if both are true. This can restated in an abbreviated form of a table:
T and T → T
T and F → T
F and T → T
F and F → F
Therefore replace the above pairs of truth values with single truth values to get the table:
1 F,
2 T,
3 T,
4 T,
5 T,
6 T,
7 T,
8 T,
9 F
Consider now x is in {2,3,4,5,6} or x is in {4,5,6,7,8} to be a single open statement. This table shows the truth set of the alternation to be {2,3,4,5,6,7,8}.
Duality between intersection and union
The following statements are equivalent to definition 2.1] and statement [2.8] concerning intersections and unions, respectively:
(*) The open statement x is in S1 ∩ S2 assigns T to an object in U if and only if both open statements x is in S1, x is in S2 assign T to that same object.
(**) The open statement x is in S1 ∪ S2 assigns F to an object in U if and only if both open statements x is in S1, x is in S2 assign F to that same object.
The statements are the same except for the colored symbols. If the symbols ∩ and T are replaced by ∪ and F then statement (*) becomes sstatement (**). And conversely, replacing blue symbols by red symbols. (*) and (**) are dual statements. Intuitively speaking, the operations of intersection and union are duals of each other. Such a replacement in a discussion about intersections may produce a corresponding discussion about unions, and conversely.
Five relative positions of two sets
The five relative positions of one set and another set are shown above:
disjoint (Fig A), intersecting (Fig B), inclusion (Fig C), equality (Fig D), complimentary (Fig E)
In Figs A,B,C,D consider set stationary, and set moving to the right.
Start with the two sets being separate (Fig A).
Then by moving far enough to the right,
set intersects with the set (Fig B).
Again moving far enough to the right,
set finds itself inside set (Fig C).
Remaining inside set, set expands until
it fills up all of set (Fig D)
Go back to Fig A. Both sets expand, without intersecting, until
together they fill up all of the universe (Fig E).
Chains of sets and chains of open statements
Let p1(x) be any open statement equivalent to x is in S1;
Let p2(x): be any open statement equivalent to x is in S2;
Let p3(x): be any open statement equivalent to x is in S3;
.............................................................................................................
Let pn(x): be any open statement equivalent to x is in Sn.
Then the chain of n included sets induces the chain
p1(x) ==> p2(x) ==> p3(x) ==> ... ==> pn(x)
of implications.
Supporting argument for direct proof through a chain of implications
[6.1] (Direct proof of a universally true implication) An implication of open statements must be universally true if there is a chain of (universally) true implications joining the hypothesis to the conclusion.
Notation: p(x) ==> q(x) is universally true if there is a chain of (universally) true implications:
p(x) ==> r1(x) ==> r2(x) ==> r3(x) ==> ... ==> rn(x) ==> q(x).
joining p(x) to q(x).
The actual proof uses mathematical induction. But here is a supporting argument.
The implication p(x) ==> q(x) is universally true if there is no counterexample to it in the universe. Assume that there is a counterexample b. Then p(b) is T and q(b) is F.
Since the implication rn(x) ==> q(x) is universally true, then b cannot be a counterexample to it, that is, rn (b) be T and q(b) be F. Therefore rn(b) must be F.
Since the implication rn-1(x) ==>rn(x) is universally true, then b cannot be a counterexample to it, that is, rn-1 (b) be T and rn(x) be F. Therefore rn-1(b) must be F.
..... This continues going backwards on the chain of implications until the first implication ....
Since the implication p(x) ==> r1(x) is universally true, then b cannot be a counterexample to it, that is, p(b) be T and r1(x) be F. Therefore p(b) must be F.
But by assumption, p(b) was T. This contradiction proves the assumption that there is a counterexample to the implication p(x) ==> q(x) is false