Here is the truth table for an implication:
$$ \begin{array}{ccccc}
P & & Q & & P \to Q \\
T & & T & & T \\
T & & F & & F \\
F & & T & & T \\
F & & F & & T \end{array} $$
You can think of an implication as a conditional promise. If you keep the promise, it's true. If you break the promise, it's false.
If I tell my kids, "I'll give you a cookie if you clean up." Then they clean up. I better give them a cookie. If I don't, I've lied. However, if they don't clean up, I can either give them a cookie or not. I didn't promise either if they didn't keep up their end of the bargain.
So in other words, an implication is false only if the hypothesis is true and conclusion is false.
Logically $P \to Q$ is equivalent to $\neg P \vee Q$
I had a computer science professor who was fond of promising his kids things given a false premise. This way he wasn't compelled to follow through. Example: "If the moon is made of green cheese, I'll give you an x-box."
$$
\newcommand{\T} {\color{blue}{\text{T}}}
\newcommand{\F} {\color{red}{\text{F}}}
\newcommand{\?} {\color{green}{\text{?}}}$$
This is one way of writing truth tables that comes out pretty nice. Start by writing out the 4 cases for $p$ and $q$:
$$\begin{array} {cccccccccccc}
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p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\
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\T & & && & \T & & \T & & \T & & \T \\
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\T & & && & \T & & \F & & \F & & \T \\
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\F & & && & \F & & \T & & \T & & \F \\
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\F & & && & \F & & \F & & \F & & \F \\
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\end{array}$$
Then start filling in the table 1 operator at a time, starting with the first operator evaluated, the $\lnot$ in the $\lnot p$:
$$\begin{array} {cccc|c|ccccccc}
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p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\
%
\T & & && \F & \T & & \T & & \T & & \T \\
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\T & & && \F & \T & & \F & & \F & & \T \\
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\F & & && \T & \F & & \T & & \T & & \F \\
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\F & & && \T & \F & & \F & & \F & & \F \\
%
\end{array}$$
Then the value of the $\lor$ in $\lnot p \lor q$, using the values in the $\lnot$ column and the $q$ column:
$$\begin{array} {cccccc|c|ccccc}
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p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\
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\T & & && \F & \T & \T & \T & & \T & & \T \\
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\T & & && \F & \T & \F & \F & & \F & & \T \\
%
\F & & && \T & \F & \T & \T & & \T & & \F \\
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\F & & && \T & \F & \T & \F & & \F & & \F \\
%
\end{array}$$
Continue filling in for $\implies$ next, then $\land$, then $\lnot$, then $\lor$:
$$\begin{array} {c|c|cccccccccc}
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p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\
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\T & \? & \? && \F & \T & \T & \T & \? & \T & \? & \T \\
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\T & \? & \? && \F & \T & \F & \F & \? & \F & \? & \T \\
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\F & \? & \? && \T & \F & \T & \T & \? & \T & \? & \F \\
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\F & \? & \? && \T & \F & \T & \F & \? & \F & \? & \F \\
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\end{array}$$
If all the values in the final column are true, then the statement is a tautology. If all the values in the final column are false, then it is a contradiction. Otherwise, it could be either (depends on the values of $p$ and $q$, not quite right to say neither).
Best Answer
In the context of propositional logic, a model is simply an assignment of truth values to the basic propositional variables. The statement $p\land q$ is true in any model that satisfies both $p$ and $q$, i.e., in which $p$ and $q$ are both true; the statement $p\lor q$, on the other hand, is true in any model that satisfies at least one of $p$ and $q$. This information is precisely what you’re collecting when you write out a truth table.
To see whether a formula $F_2$ is a logical consequence of a formula $F_1$, write out their joint truth table. Find the rows in which $F_1$ is true; those are essentially the models that satisfy $F_1$. Check to see whether $F_2$ is also true in all of those rows/models; if it is, then every model that satisfies $F_1$ automatically satisfies $F_2$, and $F_2$ is therefore a logical consequence of $F_1$. Here’s an example.
$$\begin{align*} &F_1:\quad\big(p\to(q\lor r)\big)\land\lnot r\\ &F_2:\quad p\to q \end{align*}$$
Here’s the combined truth table:
$$\begin{array}{c|c} p&q&r&q\lor r&p\to(q\lor r)&\lnot r&p\to(q\lor r)\big)\land\lnot r&p\to q\\ \hline T&T&T&T&T&F&F&\color{blue}{T}\\ \color{green}{T}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\ T&F&T&T&T&F&F&F\\ T&F&F&F&F&T&F&F\\ F&T&T&T&T&F&F&\color{blue}{T}\\ \color{green}{F}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\ F&F&T&T&T&F&F&\color{blue}{T}\\ \color{green}{F}&\color{green}{F}&\color{green}{F}&F&T&T&\color{red}{T}&\color{blue}{T} \end{array}$$
The $T$’s for $F_1$ are in red, and those for $F_2$ are in blue. As you can see, in every row in which $F_1$ is true, $F_2$ is also true; the truth values of $p,q$, and $r$ in those rows are in green. $F_1$ holds in precisely those models in which $p$ and $q$ are true and $r$ is false, or $q$ is true and $p$ and $r$ are false, or all three are false. And in all such models $F_2$ holds as well, so $F_2$ is a logical consequence of $F_1$.
Note, though, that $F_2$ holds in some models in which $F_1$ does not hold, e.g., those in which $p,q$, and $r$ are all true. Thus, $F_1$ is not a logical consequence of $F_2$, and therefore $F_1$ and $F_2$ are not logically equivalent.