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Use the truth tables for the \"and\" operator $\\wedge$ and the implication $\\Rightarrow$. Remember that $p\\Rightarrow q$ is true whenever $p$ is false or $q$ is true.

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Complete the following truth table; enter $0$ for false and $1$ for true.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$p$$q$$r$$q\\land r$$p\\Rightarrow(q\\land r)$$p\\Rightarrow q$$p\\Rightarrow r$$(p\\Rightarrow q)\\land(p\\Rightarrow r)$
000[[0]][[8]][[16]][[24]][[32]]
001[[1]][[9]][[17]][[25]][[33]]
010[[2]][[10]][[18]][[26]][[34]]
011[[3]][[11]][[19]][[27]][[35]]
100[[4]][[12]][[20]][[28]][[36]]
101[[5]][[13]][[21]][[29]][[37]]
110[[6]][[14]][[22]][[30]][[38]]
111[[7]][[15]][[23]][[31]][[39]]
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From this truth table, are the expressions $p\\Rightarrow(q\\land r)$ and $(p\\Rightarrow q)\\land (p\\Rightarrow r)$ logically equivalent?

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Yes

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No

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In this question you are asked to construct a truth table for the expressions

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$p\\Rightarrow(q\\wedge r)$ and $(p\\Rightarrow q)\\wedge (p\\Rightarrow r)$, using all possible combinations of truth values for $p,q,r$.

\n

Recall that $0$ means false and $1$ means true.

\n

\n

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Create a truth table for a logical expression of the form $(a \\operatorname{op1} b) \\operatorname{op2}(c \\operatorname{op3} d)$ where $a, \\;b,\\;c,\\;d$ can be the Boolean variables $p,\\;q,\\;\\neg p,\\;\\neg q$ and each of $\\operatorname{op1},\\;\\operatorname{op2},\\;\\operatorname{op3}$ one of $\\lor,\\;\\land,\\;\\to$.

\n

For example: $(p \\lor \\neg q) \\land(q \\to \\neg p)$.

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