Help, find a statement using a truth table

  • Thread starter Thread starter jhlee127
  • Start date Start date
  • Tags Tags
    Table Truth table
AI Thread Summary
To express the statement S from the provided truth table, one can create a disjunction of all cases where S is true. This involves combining conditions using logical operators like "and" and "not" for each true instance of S. Simplifying the resulting expression can help clarify the symbolic representation of S. The discussion also touches on the context of studying logic, whether for mathematics or engineering applications. Understanding how to derive and simplify logical statements is essential for both fields.
jhlee127
Messages
1
Reaction score
0
i don't understand how i can find a statement in given truth table.

p | q | r | S |
==========
T | T | T | F|
T |T | F | T |
T | F | T | T |
T | F | F | F |
F | T | T | T |
F | T | F | F |
F | F | T | T |
F | F | F | F |


please help me.
 
Physics news on Phys.org
jhlee127 said:
i don't understand how i can find a statement in given truth table.
I assume you mean "I don't understand how to express the statement S in symbolic form when given the following truth table."

If you can't make a clever guess for S, you could simply write it as a big "or" statement containing all the cases when it is true:

S \equiv (p \land q \land \lnot r) \lor (p \land \lnot q \land r) \lor ... etc.

then try to simplify the symbolic statement.

Are you studying logic in course on mathematics or are you studying logic in an engineering course in order to design logic circuits?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top