Section gluD.
Hypothesis j1 : IN (PAIR X Z) G.
Hypothesis i1 : IN (PAIR X Y) G.
Proof step4 X Y Z i1 j1.
End gluD.
Let step6 : IN (PAIR X Y) F -> EQ Y Z.
Proof
union_th4 (IN (PAIR X Y) F -> EQ Y Z) F G (PAIR X Z) j gluA_Out gluB_Out.
Let step7 : IN (PAIR X Y) G -> EQ Y Z.
Proof
union_th4 (IN (PAIR X Y) G -> EQ Y Z) F G (PAIR X Z) j gluC_Out gluD_Out.

Lemma glu_wd_Out : EQ Y Z.