Section 2.4: Problem 11 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

(Eq3) Show that $\vdash\forall x\forall y\forall z(x=y\rightarrow y=z\rightarrow x=z).$

By the Generalization Theorem, it is sufficient to show $\vdash x=y\rightarrow y=z\rightarrow x=z$ . By axiom group 5, we have $\vdash x=x$ . By axiom group 6, we have $\vdash x=y\rightarrow x=x\rightarrow y=x$ and $\vdash y=x\rightarrow y=z\rightarrow x=z$ . The first two, by Rule T, imply $\vdash x=y\rightarrow y=x$ , which together with the third one, again by Rule T, imply $\vdash x=y\rightarrow y=z\rightarrow x=z$ .

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Section 2.4: Problem 11 Solution

Section 2.4: Problem 11 Solution