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Seems a distributive lattice is a semiring; if a semifield is a semiring where multiplication by a non-zero $x$ is invertible, then $x \wedge x = x \wedge \top$ implies $x = \top$ assuming embeddability in a semifield. So, the answer is no.
Connes and Consani, for instance, say a semifield is a semiring (ie a rig) where the nonzero elements are all invertible.
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