Question #216850

Using laws of logic solve the following compound propositions. Also indicate the names

of laws.

[𝑝 ∧ (¬𝑝 ∨ 𝑞)] ∨ [𝑞 ∧ ¬(𝑝 ∧ 𝑞)]


1
Expert's answer
2021-07-14T09:37:56-0400

Let us use laws of logic solve the following compound propositions.


[𝑝(¬𝑝𝑞)][𝑞¬(𝑝𝑞)][𝑝 ∧ (¬𝑝 ∨ 𝑞)] ∨ [𝑞 ∧ ¬(𝑝 ∧ 𝑞)]


| use the distributive law |


=[(𝑝¬𝑝)(p𝑞)][𝑞¬(𝑝𝑞)]=[(𝑝 ∧ ¬𝑝) ∨(p\land 𝑞)] ∨ [𝑞 ∧ ¬(𝑝 ∧ 𝑞)]


| use de Morgan law |


=[(𝑝¬𝑝)(p𝑞)][𝑞(¬𝑝¬𝑞)]=[(𝑝 ∧ ¬𝑝) ∨(p\land 𝑞)] ∨ [𝑞 ∧ (¬𝑝 \lor\neg 𝑞)]


| use the law of contradiction |


=[F(p𝑞)][𝑞(¬𝑝¬𝑞)]=[F ∨(p\land 𝑞)] ∨ [𝑞 ∧ (¬𝑝 \lor\neg 𝑞)]


| use the distributive law |


=[F(p𝑞)][(𝑞¬𝑝)(q¬𝑞)]=[F ∨(p\land 𝑞)] ∨ [(𝑞 ∧ ¬𝑝) \lor(q\land \neg 𝑞)]


| use the constant law |


=(p𝑞)[(𝑞¬𝑝)(q¬𝑞)]=(p\land 𝑞) ∨ [(𝑞 ∧ ¬𝑝) \lor(q\land \neg 𝑞)]


| use the law of contradiction |


=(p𝑞)[(𝑞¬𝑝)F]=(p\land 𝑞) ∨ [(𝑞 ∧ ¬𝑝) \lor F]


| use the constant law |


=(p𝑞)(𝑞¬𝑝)=(p\land 𝑞) ∨ (𝑞 ∧ ¬𝑝)


| use the commutative law |


=(qp)(𝑞¬𝑝)=(q\land p) ∨ (𝑞 ∧ ¬𝑝)


| use the distributive law |


=q(p¬𝑝)=q\land (p ∨ ¬𝑝)


| use the law of the excluded third |


=qT=q\land T


| use the constant law |


=q=q



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