~(~Pv(Q^R))<->(S->P)
Will it run?
Will it run?
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Topic Starter
https://en.m.wikipedia.org/wiki/Truth_tableCorne2Plum3 wrote:
idk? https://en.wikipedia.org/wiki/Semantics_(computer_science)
https://en.m.wikipedia.org/wiki/Natural_deduction
These Wikipedia articles suck, it makes it look much more complex that it actually is
Sorry, I don't speak programming. but I asked ChatGPT-4, although I cannot guarantee that it is correct...
Topic Starter
It is indeed correct, although without context it means nothingBehrauder wrote:
Sorry, I don't speak programming. but I asked ChatGPT-4, although I cannot guarantee that it is correct...
he given logical expression is:
[ \neg (\neg P \lor (Q \land R)) \leftrightarrow (S \rightarrow P) ]
Let’s break it down step by step:
Negation of the Left Side:
(\neg (\neg P \lor (Q \land R)))
Apply De Morgan’s law: (\neg (\neg P) \land \neg (Q \land R))
Simplify: (P \land (\neg Q \lor \neg R))
Implication on the Right Side:
(S \rightarrow P)
This means: “If (S) is true, then (P) must be true.”
Equivalence:
The entire expression is an equivalence, so both sides must be equivalent.
Therefore, we need to check if (P \land (\neg Q \lor \neg R)) is equivalent to (S \rightarrow P).
Truth Table:
Let’s create a truth table for both sides and compare the results.
Table
(P) (Q) (R) (S) (P \land (\neg Q \lor \neg R)) (S \rightarrow P)
T T T T T T
T T F T T T
T F T T T T
T F F T T T
F T T T F F
F T F T F F
F F T T F F
F F F T F F
Conclusion:
Both sides have the same truth values for all possible combinations of (P), (Q), (R), and (S).
Therefore, the given logical expression is true.
[ \neg (\neg P \lor (Q \land R)) \leftrightarrow (S \rightarrow P) ]
Let’s break it down step by step:
Negation of the Left Side:
(\neg (\neg P \lor (Q \land R)))
Apply De Morgan’s law: (\neg (\neg P) \land \neg (Q \land R))
Simplify: (P \land (\neg Q \lor \neg R))
Implication on the Right Side:
(S \rightarrow P)
This means: “If (S) is true, then (P) must be true.”
Equivalence:
The entire expression is an equivalence, so both sides must be equivalent.
Therefore, we need to check if (P \land (\neg Q \lor \neg R)) is equivalent to (S \rightarrow P).
Truth Table:
Let’s create a truth table for both sides and compare the results.
Table
(P) (Q) (R) (S) (P \land (\neg Q \lor \neg R)) (S \rightarrow P)
T T T T T T
T T F T T T
T F T T T T
T F F T T T
F T T T F F
F T F T F F
F F T T F F
F F F T F F
Conclusion:
Both sides have the same truth values for all possible combinations of (P), (Q), (R), and (S).
Therefore, the given logical expression is true.