I'm considering the following problem from Elliott Mendelson's book "Introduction to Mathematical Logic".
The proof of (b) he gives is as follows: A conjunction $\mathscr{E}$ of the form $B_1^{*}\wedge B_2^{*}\dots\wedge B_n^{*},$ where each $B_i^{*}$ is either $B_i$ or $\neg B_i,$ is said to be eligible if some assignment of truth values to the statement letters of $\mathscr{B}$ that makes $\mathscr{B}$ true also makes $\mathscr{E}$ true. Let $\mathscr{C}$ be the disjunction of all eligible conjunctions. $\square$
In the particular case of (c), this leads us to the statement form $\mathscr{C}=(B_1\Rightarrow B_2),$ and I agree both $\mathscr{B}\Rightarrow\mathscr{C}$ and $\mathscr{C}\Rightarrow\mathscr{D}$ are tautologies.
However, I'm concerned about what happens if we change $\mathscr{D}$ to be $$((B_1\wedge C)\Rightarrow(B_2\wedge C))\wedge(B_1\vee B_2\vee \neg C)$$
Essentially this changes the truth table of $\mathscr{D}$ to give $F$ when $B_1,B_2$ are $F$ and $C$ is $T$.
It's certainly still true that $\mathscr{B}\Rightarrow\mathscr{D}$ is a tautology, and applying the algorithm in the solution still suggests that $\mathscr{C}=(B_1\Rightarrow B_2)$ should do the job.
However, note that we now have a case where $\mathscr{C}$ is $T$ yet $\mathscr{D}$ is $F$ - namely when $B_1,B_2$ are $F$ and $C$ is $T$. Thus we no longer have that $\mathscr{C}\Rightarrow \mathscr{D}$ is a tautology.
In fact, for the altered example, I'm fairly confident no suitable $\mathscr{C}$ exists, as its value when $B_1,B_2$ are $F$ has to both be $T$ (since $\mathscr{B}\Rightarrow\mathscr{C}$) and $F$ (since $\mathscr{C}\Rightarrow\mathscr{D}$).
Where exactly am I going wrong?
