There are some problems with my previous post, sorry
(1) q -> fp = f(q -> p) ; in particular, f is inflationary. This has been pointed out by Ingo Blechschmidt
Actually inflationarity does not follow from this directly, (1) only gives q -> fq = f1 To show f1 = 1 use f(fp -> p) with p = 1
(7) q -> fp = q -> f(q & p): by (1) both are equal to f(q -> p) = f(q -> (q & p))
(8) fq & fp -> f(q & p): in (7), take fq & fp in place of q, then the lhs of (7) becomes true
lhs indeed becomes true but the rhs is not yet what we want, it is (fq & fp) -> f(fq & fp & p) = f(fq & p). To finalize, f(fq & p) -> f(q & p) has to be proved, and here it seems I cannot get away without another use of monotonicity: from fq & p -> f(q & p) it gives f(fq & p) -> ff(q & p) and then one can use (6). As for fq & p -> f(q & p), it is obtained from (7) as explained in the last remark of my previous post. Mamuka [For admin and other information see: http://www.mta.ca/~cat-dist/ ]