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

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SUMMARY

The discussion focuses on proving that the expression \(\left[\left(p\,\longrightarrow\,q\right)\,\wedge\,\left(q\,\longrightarrow\,r\right)\right]\,\longrightarrow\,\left(p\,\longrightarrow\,r\right)\) is a tautology. Participants suggest using logical equivalences and truth tables to demonstrate this. The approach involves expanding implications and simplifying the expression to show that it holds true under all interpretations. The consensus emphasizes the importance of symbolic manipulation and truth table construction for clarity in logical proofs.

PREREQUISITES
  • Understanding of logical implications and equivalences
  • Familiarity with truth tables in propositional logic
  • Basic knowledge of symbolic logic notation
  • Experience with logical proofs and tautologies
NEXT STEPS
  • Learn how to construct and analyze truth tables for logical expressions
  • Study logical equivalences and their applications in proofs
  • Explore methods for symbolic manipulation in propositional logic
  • Practice proving tautologies using different logical frameworks
USEFUL FOR

Students of mathematics, particularly those studying logic and proofs, as well as educators teaching propositional logic concepts.

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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}<br /> \left[ {\left( {p\; \to \;q} \right)\; \wedge \;\left( {q\; \to \;r} \right)} \right]\; \to \;\left( {p\; \to \;r} \right) \\ <br /> \left[ {\left( {\neg p\; \vee \;q} \right)\; \wedge \;\left( {\neg q\; \vee \;r} \right)} \right]\; \to \;\left( {p\; \to \;r} \right) \\ <br /> \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) \\ <br /> \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) \\ <br /> \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) \\ <br /> \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) \\ <br /> \end{array}

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

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