Question

Prove the following theorems: $\forall p_{o \beta} \forall q_{\mathrm{o} \beta}=E_{\mathrm{o}(\mathrm{o} \beta)(\mathrm{o} \beta)} p_{\mathrm{o} \beta} q_{\mathrm{o} \beta}=\cdot \overline{\overline{p_{\mathrm{o} \beta}}}=\overline{\overline{q_{\mathrm{o} \beta}}}$

   Prove the following theorems:
$\forall p_{o \beta} \forall q_{\mathrm{o} \beta}=E_{\mathrm{o}(\mathrm{o} \beta)(\mathrm{o} \beta)} p_{\mathrm{o} \beta} q_{\mathrm{o} \beta}=\cdot \overline{\overline{p_{\mathrm{o} \beta}}}=\overline{\overline{q_{\mathrm{o} \beta}}}$
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An Introduction to Mathematical Logic and Type Theory: To Truth Through Proof (Computer Science & Applied Mathematics)
An Introduction to Mathematical Logic and Type Theory: To Truth Through Proof (Computer Science & Applied Mathematics)
Peter B. Andrews 1st Edition
Chapter 6, Problem 4 ↓

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The notation $\forall p_{o \beta} \forall q_{\mathrm{o} \beta}=E_{\mathrm{o}(\mathrm{o} \beta)(\mathrm{o} \beta)} p_{\mathrm{o} \beta} q_{\mathrm{o} \beta}=\cdot \overline{\overline{p_{\mathrm{o} \beta}}}=\overline{\overline{q_{\mathrm{o} \beta}}}$ appears to be  Show more…

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Prove the following theorems: $\forall p_{o \beta} \forall q_{\mathrm{o} \beta}=E_{\mathrm{o}(\mathrm{o} \beta)(\mathrm{o} \beta)} p_{\mathrm{o} \beta} q_{\mathrm{o} \beta}=\cdot \overline{\overline{p_{\mathrm{o} \beta}}}=\overline{\overline{q_{\mathrm{o} \beta}}}$
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