Logic: Proving tautological consequence I'm having trouble proving this tautological consequence. I'd hope that you guys can maybe oversee my process and identify errors, because I went over this couple of times and I arrive at the same conclusion.
The question goes like this:

$A \Rightarrow  B$
$C \Rightarrow B$
Therefore: $(A \lor C) \Rightarrow B$
Q: Show that the conclusion of the arguments is a tautological
  consequence of the premises using truth tables.

This is how I tried to solve it:
My solution to the problem. It seems like I'm missing something very fundamental here.

The problem is really:
A implies B
C implies B is
(A implies B) and (C implies B)
And then it should be easily provable
edit: 
\begin{array}{ccc|c@{}ccc@{}ccccc@{}ccc@{}ccc@{}ccc@{}c}
a&b&c&(&a&\lor&c&)&\rightarrow&b&\leftrightarrow&(&a&\rightarrow&b&)&\land&(&c&\rightarrow&b&)\\\hline
1&1&1&&1&1&1&&1&1&\mathbf{1}&&1&1&1&&1&&1&1&1&\\
1&1&0&&1&1&0&&1&1&\mathbf{1}&&1&1&1&&1&&0&1&1&\\
1&0&1&&1&1&1&&0&0&\mathbf{1}&&1&0&0&&0&&1&0&0&\\
1&0&0&&1&1&0&&0&0&\mathbf{1}&&1&0&0&&0&&0&1&0&\\
0&1&1&&0&1&1&&1&1&\mathbf{1}&&0&1&1&&1&&1&1&1&\\
0&1&0&&0&0&0&&1&1&\mathbf{1}&&0&1&1&&1&&0&1&1&\\
0&0&1&&0&1&1&&0&0&\mathbf{1}&&0&1&0&&0&&1&0&0&\\
0&0&0&&0&0&0&&1&0&\mathbf{1}&&0&1&0&&1&&0&1&0&
\end{array}
