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Because of ~(G0 /\ G1) iff ~G0 \/ ~G1, the indirect method leads to
~~>
K G0 /\ G1 ~~>
K union {~G0 \/ ~G1} F(alse)
K union {~G0} F(alse)
K union {~G1} F(alse)
We have to prove G0 /\ G1.
We will see later this technique of "case distinction".