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The statement,” Every comedian is funny” where C(x) is “x is a comedian” and F (x) is “x is funny” and the domain consists of all people. ans. ∃x(C(x) ∧ F (x)) ∀x(C(x) ∧ F (x)) ∀x(C(x) → F (x)) ∃x(C(x) → F (x))

Question

The statement,” Every comedian is funny” where C(x) is “x is a comedian” and F (x) is “x is funny” and the domain consists of all people. ans. ∃x(C(x) ∧ F (x))

∀x(C(x) ∧ F (x))

∀x(C(x) → F (x))

∃x(C(x) → F (x))

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Solution

The correct translation of the statement "Every comedian is funny" in the language of logic is:

∀x(C(x) → F (x))

This statement is read as "For all x, if x is a comedian then x is funny". This correctly captures the meaning of the original statement, as it asserts that being a comedian (C(x)) implies being funny (F(x)) for every person x in the domain.

The other statements have different meanings:

∃x(C(x) ∧ F (x)) is read as "There exists an x such that x is a comedian and x is funny". This statement asserts that there is at least one person who is both a comedian and funny, but it does not claim this for all comedians.

∀x(C(x) ∧ F (x)) is read as "For all x, x is a comedian and x is funny". This statement asserts that every person is both a comedian and funny, which is not what the original statement says.

∃x(C(x) → F (x)) is read as "There exists an x such that if x is a comedian then x is funny". This statement asserts that there is at least one person for whom being a comedian implies being funny, but it does not claim this for all comedians.

This problem has been solved

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s this ∃xQ(x) a tautology

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ind a simplified expression for( fg)(x), and find its domain.

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