A truth table needed?

Logic Level 3

∀ a ∀ b ∀ c ( ¬ [ ( ¬ a ∧ ¬ b ) ∨ ( ¬ a ∧ c ) ∨ ( b ∧ c ) ] ∨ ( ¬ a ∨ c ) ) \large \forall a \forall b \forall c \left( \neg{\left[ (\neg{a} \land \neg{b}) \lor (\neg{a}\land c) \lor (b \land c) \right]} \lor (\neg{a} \lor c) \right)

Is the above proposition true?

False True It depends Not enough information

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Paulo Guilherme Santos by Paulo Guilherme Santos , 25, Portugal

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1 solution

We know that ∀ a ( ∀ b ( ∀ c ( ( ( a → b ) ∧ ( b → c ) ) → ( a → c ) ) ) ) \forall a \left( \forall b \left( \forall c \left(\left( \left(a \rightarrow b \right)\land \left(b \rightarrow c\right) \right) \rightarrow \left(a\rightarrow c \right) \right) \right) \right) . But we have that ∀ a ( ∀ b ( ∀ c ( ( ( a → b ) ∧ ( b → c ) ) → ( a → c ) ) ) ) ⟺ \forall a \left( \forall b \left( \forall c \left(\left( \left(a \rightarrow b \right)\land \left(b \rightarrow c\right) \right) \rightarrow \left(a\rightarrow c \right) \right) \right) \right)\iff ∀ a ( ∀ b ( ∀ c ( ( ( ¬ a ∨ b ) ∧ ( ¬ b ∨ c ) ) → ( ¬ a ∨ c ) ) ) ) ⟺ \forall a \left( \forall b \left( \forall c \left(\left( \left(\neg{a} \lor b \right)\land \left(\neg{b} \lor c \right) \right) \rightarrow \left(\neg{a} \lor c \right) \right) \right) \right) \iff ∀ a ( ∀ b ( ∀ c ( ¬ ( ( ¬ a ∨ b ) ∧ ( ¬ b ∨ c ) ) ∨ ( ¬ a ∨ c ) ) ) ) ⟺ \forall a \left( \forall b \left( \forall c \left(\neg{\left( \left(\neg{a} \lor b \right)\land \left(\neg{b} \lor c \right) \right)} \lor \left(\neg{a} \lor c \right) \right) \right) \right) \iff ∀ a ( ∀ b ( ∀ c ( ¬ ( ( ¬ a ∧ ¬ b ) ∨ ( ¬ a ∧ c ) ∨ ( b ∧ ¬ b ) ∨ ( b ∧ c ) ) ∨ ( ¬ a ∨ c ) ) ) ) ⟺ \forall a \left( \forall b \left( \forall c \left(\neg{\left( \left(\neg{a} \land \neg{b} \right)\lor \left(\neg{a} \land c \right) \lor \left(b \land \neg{b} \right) \lor \left(b \land c \right)\right)} \lor \left(\neg{a} \lor c \right) \right) \right) \right) \iff ∀ a ( ∀ b ( ∀ c ( ¬ ( ( ¬ a ∧ ¬ b ) ∨ ( ¬ a ∧ c ) ∨ ( b ∧ c ) ) ∨ ( ¬ a ∨ c ) ) ) ) \forall a \left( \forall b \left( \forall c \left(\neg{\left( \left(\neg{a} \land \neg{b} \right)\lor \left(\neg{a} \land c \right) \lor \left(b \land c \right)\right)} \lor \left(\neg{a} \lor c \right) \right) \right) \right)

As wanted.

1 Helpful 0 Interesting 0 Brilliant 1 Confused

By Boolean algebra, I get a + b + c + 1 a+b+c+1 . I forgot that a + 1 = 1 a+1 = 1 :(

Jake Lai - 5 years, 11 months ago

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