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Proof step 1 from proof isEqRel-reflexive (previous step)

⊦ isEqRel(eqstate) ↔ (∀ s. eqstate(s, s)) ∧ (∀ s, s0. eqstate(s, s0) → eqstate(s0, s)) ∧ (∀ s, s0, s1. eqstate(s, s0) ∧ eqstate(s0, s1) → eqstate(s, s1))

Lemma isEqRel-def