Hi Erick, I don’t understand the need for the axiom because the corresponding theorem is proved by TLAPS: CONSTANT E(_) THEOREM Leibniz == \A p,q : (p = q) => (E(p) \equiv E(q)) OBVIOUS Similarly, p && q == p /\ q THEOREM Thm3_84a == ASSUME NEW e, NEW f PROVE (e = f) && E(e) \equiv (e = f) && E(f) <1>1 ((e = f) && E(e) \equiv (e = f) && E(f)) \equiv ((e = f) && (E(e) \equiv E(f)) \equiv (e = f)) (* <2> DEFINE t == (e = f) <2> DEFINE u == E(e) <2> DEFINE v == E(f) <2> HIDE DEF u,v,t <2>1 t && (u \equiv v) \equiv (t && u \equiv (t && v \equiv t)) BY Thm3_49 <2>2 (t && (u \equiv v) \equiv (t && u \equiv (t && v \equiv t))) \equiv ((t && u \equiv t && v) \equiv (t && (u \equiv v) \equiv t)) BY EqAssoc <2> QED BY <2>1, <2>2 DEF t, u, v *) <1>2 @ \equiv ((e = f) => (E(e) \equiv E(f))) \* BY ImplDef3 <1> QED BY <1>1, <1>2\*, Leibniz (I elided the parts of the proof that refer to facts that come from elsewhere.) Stephan
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