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We can do a short proof here:entropy1 said:But actually, if ##A \rightarrow B## AND ##C \rightarrow NOT(B)##, then I wonder if C=True results in A=NOT True (or, of course, A=True in C=NOT True).
To be proven: ##C \rightarrow \neg A##
##1: A \rightarrow B## assumption 1
##2: C \rightarrow \neg B## assumption 2
##3: C## assumption 3
##4: \neg B## modus ponens 2,3
##5: \neg A ## modus tollens 1,4
##6: C \rightarrow \neg A## hypothethical syllogism (1,2),3,(4),5
##-## and there you have it ##\dots##
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