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KV-Diagram Sheet 09
((d|a?d:(c->a))|((a?c:c)?d<->d:!b)->c)

I have a question regarding this propositional formular. Why is the case a=1,b=0,c=0,d=0 true? In the solution it says that its true. Due to the precedence rules, 'or' is prioritized over the 'implication' (narrow), that means that either the left part of 'or' or the right part implicies c. Do i misunderstand the formular?

in * Other Teaching Fields by (410 points)

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You are right. This is a mistake. for a=1, b=0, c=0, d=0 the formula should be false. You can also double check with the tools.

 ((d|a?d:(c->a))|((a?c:c)?d<->d:!b)->c)
⇔((0|1?0:(0->1))|((1?0:0)?0<->0:!0)->0)
⇔((1?0:1)|(0?0<->0:1)->0)
⇔(0|1->0)
⇔(1->0)
⇔0
by (2.4k points)

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