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  "body": "\n## Definition\n\nLet $\\mathcal{V}$ and $\\mathcal{W}$ be [[varieties|AJNZJD8442]] of algebras.\nLet $\\rho$ be an [[interpretation|32PDB3WQ4D]] of $\\mathcal{V}$ in $\\mathcal{W}$, which sends each $A \\in \\mathcal{W}$ to an algebra $A^{\\rho} \\in \\mathcal{V}$ on the same set.\nLet $\\theta$ be an interpretation of $\\mathcal{W}$ in $\\mathcal{V}$, which sends each $B \\in \\mathcal{V}$ to an algebra $B^{\\theta} \\in \\mathcal{W}$ on the same set.\nThe pair $(\\rho, \\theta)$ is an *equivalence* if $(A^{\\rho})^{\\theta} = A$ for all $A \\in \\mathcal{W}$ and $(B^{\\theta})^{\\rho} = B$ for all $B \\in \\mathcal{V}$.\nThe varieties $\\mathcal{V}$ and $\\mathcal{W}$ are *equivalent*, or *term equivalent*, if such a pair exists.\n\n## Remarks\n\nExample: Boolean algebras and Boolean rings are equivalent.\nOne translation is $x + y \\mapsto (x \\wedge \\neg y) \\vee (\\neg x \\wedge y)$, $x \\cdot y \\mapsto x \\wedge y$, with $0$ and $1$ kept.\nThe other is $x \\vee y \\mapsto x + y + xy$, $x \\wedge y \\mapsto xy$, $\\neg x \\mapsto 1 + x$.\nGoing there and back recovers each operation, for example $x + y + xy$ becomes the join again in every Boolean algebra.\nNon-example: the variety of Boolean algebras is not equivalent to the variety of distributive lattices with signature $\\vee, \\wedge$.\nThere is not even an interpretation of Boolean algebras in distributive lattices, because every lattice term contains a variable, so no term can stand for the constant $0$.\nEquivalent varieties have, on each set, the same subalgebras, the same congruences and the same term operations.\n\n## Explanations\n\n### L1 human\n\nTwo varieties are equivalent when they describe the same structures with different but interchangeable basic operations.\n",
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