$A$$\rightarrow$$(B$ $\vee$ $C$ ) , $B$ $\rightarrow$ $C$ $\vDash$ $A$ $\rightarrow$ $D$
I think it's wrong but I have no idea how to prove.
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1I'll bet that the final letter is supposed to be "C" instead of "D". In that case the claim is true. You might want to check with the source to see if this was a typo. – John Hughes Apr 04 '16 at 12:01
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Thanks, but I've checked it and it's "D" @JohnHughes – Orange Apr 04 '16 at 12:02
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OK. It's wrong. The assignment $T, T, T, F$ to $A,B,C,D$ respectively shows this. – John Hughes Apr 04 '16 at 20:06
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It is provable that $A \to ( B \vee C ) ~,~ B \to C ~\vDash~ A \to \color{red}C$
$$\begin{array}{|l|l:l|}\hline 1 & A\to (B\vee C) & \textsf{Premise} \\ 2 & B\to C & \textsf{Premise} \\ \hdashline 3 & \quad \neg C & \textsf{Assume} \\ 4 & \quad \neg B & 2,3,\to\textsf{Elim (MP)} \\ 5 & \quad \neg(B\vee C) & 3,4,\wedge\textsf{Intro},\textsf{deMorgan's} \\ 6 & \quad \neg A & 1,5,\to\textsf{Elim (MP)} \\ \hline 7 & A\to C & 3,6,\to\textsf{Intro} \\ \hline \end{array}$$
However, those premises do not entail $A\to D$.
Graham Kemp
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thanks for your reply, but really it's $A$ $\rightarrow$ $D$ . by the truth table, when A=true,B=true,C=true,D=false, left-hand is true but right-hand is false. then this one is false. (a counter example) is this right solution? – Orange Apr 04 '16 at 12:26
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$A\implies D$ is false when $A$ is true and $D$ false. The other implications will be true when $C$ be true.
Martín-Blas Pérez Pinilla
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