Show that ((p implies q) and (q implies r)) implies (p implies r) is a tautology

[VinnyCee]

Homework Statement

Show that $$\left[\left(p\,\longrightarrow\,q\right)\,\wedge\,\left(q\,\longrightarrow\,r\right)\right]\,\longrightarrow\,\left(p\,\longrightarrow\,r\right)$$ is a tautology.

Homework Equations

Logical equivalences.

The Attempt at a Solution

$$\begin{array}{l} \left[ {\left( {p\; \to \;q} \right)\; \wedge \;\left( {q\; \to \;r} \right)} \right]\; \to \;\left( {p\; \to \;r} \right) \\ \left[ {\left( {\neg p\; \vee \;q} \right)\; \wedge \;\left( {\neg q\; \vee \;r} \right)} \right]\; \to \;\left( {p\; \to \;r} \right) \\ \left\{ {\left[ {\left( {\neg p\; \vee \;q} \right)\; \wedge \;\neg q} \right]\; \vee \;\left[ {\left( {\neg p\; \vee \;q} \right)\; \wedge \;r} \right]} \right\}\; \to \;\left( {p\; \to \;r} \right) \\ \left\{ {\left[ {\left( {\neg p\; \wedge \;\neg q} \right)\; \vee \;\left( {q\; \wedge \;\neg q} \right)} \right]\; \vee \;\left[ {\left( {\neg p\; \wedge \;r} \right)\; \vee \;\left( {q\; \wedge \;r} \right)} \right]} \right\} \to \;\left( {p\; \to \;r} \right) \\ \left\{ {\left[ {\left( {\neg p\; \wedge \;\neg q} \right)\; \vee \;{\rm F}} \right]\; \vee \;\left[ {\left( {\neg p\; \wedge \;r} \right)\; \vee \;\left( {q\; \wedge \;r} \right)} \right]} \right\}\; \to \;\left( {p\; \to \;r} \right) \\ \left\{ {\left[ {\neg p\; \wedge \;\neg q} \right]\; \vee \;\left[ {\left( {\neg p\; \wedge \;r} \right)\; \vee \;\left( {q\; \wedge \;r} \right)} \right]} \right\}\; \to \;\left( {p\; \to \;r} \right) \\ \end{array}$$

What now?

 

[SiddharthM]

Without any prior assumptions we need to assume (p->q) and (q->r) and from there show that p imples r. This may not be legit if your instructor wants a symbolic elimination of the "fluff". Symbollically: keep on working, you are no the right track - expand and cancel falsehoods or tautologies like you have been doing.

 

[matt grime]

As SiddharthM says, you should just expand all you implications (there are two left) as not ors. Or write out a truth table.