MHB Translate argument into symbolic form

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The discussion focuses on translating two statements about the status of a human fetus into symbolic logic. The first statement, "If we are less than certain the human fetus is a person, then we must give it the benefit of the doubt," is represented as ¬S → B. The second statement, "If we are certain the human fetus is a person, then we must accord it the right to live," is expressed as S → L. The complexity arises from the phrasing "less than certain" and "certain." The participants seek clarity on these translations and their implications in the context of the debate.
natalie206
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Hi guys,

maybe you have any idea how to translate this two statements:

If we are less than certain the human fetus is a person, then we must give it the benefit of the doubt. If we are certain the human fetus is a person, then we must accord it the right to live.
 
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Let $S$ denote "we are certain the human fetus is a person", $B$ denote "we must give it the benefit of the doubt" and $L$ denote "we must accord it the right to live". Then the first sentence says $\neg S\to B$ and the second one says $S\to L$.

Also see an important remark https://driven2services.com/staging/mh/index.php?posts/101747/.
 
Thanks! I thought it's more complicated, because of "less than certain" and "certain".

Thank you!
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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