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The equational theories of type F over X form an algebraic lattice which is isomorphic to the lattice of fully invarient congruences on T(X).

I need the proof of above theorem which is in page 103 (Corollary14.10) of Universal Algebra sankappanavar.

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By definition, $\Sigma\subseteq\operatorname{Id}(X)$ is an equational theory (of type $\mathscr{F}$) over $X$ iff $\Sigma=\operatorname{Id}_K(X)$ for some class $K$ of algebras; by Theorem 14.8 this is the case iff $\tau(\Sigma)$ is a fully invariant congruence on $\mathbf{T}(X)$, the term algebra (of type $\mathscr{F}$) over $X$, in which case we know from the proof of Theorem 14.8 we know that we can take $K=\{\mathbf{T}(X)/\tau(\Sigma)\}$.

Since $p\approx q\in\Sigma$ iff $\langle p,q\rangle\in\tau(\Sigma)$, it’s clear that the map $\Sigma\mapsto\tau(\Sigma)$ is a lattice isomorphism, so it only remains to check that the lattice of fully invariant congruences on $\mathbf{T}(X)$ is an algebraic lattice. This follows from Theorems I.5.5 and II.14.4. By 14.4 $\Theta_{\text{FI}}$ is an algebraic closure operator on $T(X)\times T(X)$, so by 5.5 the subsets of $T(X)\times T(X)$ closed under $\Theta_{\text{FI}}$ form an algebraic lattice. But the subsets of $T(X)\times T(X)$ closed under $\Theta_{\text{FI}}$ are precisely the fully invariant congruences on $\mathbf{T}(X)$.

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thanks alot my friend. It's really my pleasure to have a friend like you. – MohammadSadegh YazdanParast Dec 16 '12 at 6:11
@MohammadSadegh: It’s my pleasure to help. – Brian M. Scott Dec 16 '12 at 18:14

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