Right!
{(Mu
v
No)
[(Hi
^
Gu)
(Mu
v
No)]}
{(Mu
v
No)
[(Hi
^
Gu)
(Mu
v
No)]}
IS
an instance of theorem 1 (P
P) with "(Mu
v
No)
[(Hi
^
Gu)
(Mu
v
No)]
" substituted for "P" in both cases.
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[(Hi ^ Gu) (Mu v No)]}{(Mu v No)[(Hi ^ Gu) (Mu v No)]} P) with "(Mu v No)[(Hi ^ Gu) (Mu v No)]" substituted for "P" in both cases.