Recent content by jose diez

yes

Sorry there was a typo in (**) Corrected: (**) if A≈B & C≈D & A∩C≈B∩D then A∪C≈B∪D

This set-theory theorem is very easy to prove: (*) if A≈B & C≈D & A∩C=∅ & B∩D=∅ then A∪C≈B∪D It seems intuitive that if one replaces the strong A∩C=∅ & B∩D=∅ condition by the weaker A∩C≈B∩D the implication (**) if A≈B & C≈D & A∩C≈B∪∩D then A∪C≈BD still holds. (**) does not seem to be much...