Emma K.

# Please solve this proof ASAP.

1. (J v A) > [(S v K) > (~I • Y)]
2. (~I v ~M) > E

// J > (S > E)

By:

Emma K.

Where do the conditional and disjunction introductions come from? What rules are those?
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04/18/19

Emma K.

The rules I am to use are Modus Ponens, Modus Tolens, Hypothetical Syllagism, Disjunctive Syllagism, Simplification, Conjunction, Add, Constructive Dilemma, Double Negation, DMorgans, Commutative, Associative, Distribution, Transposition, Implication, Exportation, Tautology, and Equivalence.
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04/18/19

Hien B.

tutor
Disjunction Introduction is another name for Addition. Sometimes Conditional Introduction is also called Conditional Proof. Is that a rule you can use?
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04/18/19

Emma K.

We cannot use conditional proof.
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04/18/19

Emma K.

Also, where does conjunction and disjunction elimination come from?
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04/18/19

Hien B.

tutor
Really? I don't think this proof could be solved without using conditional proof or maybe reductio ad absurdum
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04/18/19

Hien B.

tutor
Oh sorry, that was a mistake. It should be disjunction introduction instead of elimination. Conjunction elimination is the same as Simplification
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04/18/19

Emma K.

Could you solve the proof again using the rules that I listed, so that it is easier for me to read and understand?
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04/18/19

Hien B.

tutor
With or without conditional proof?
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04/18/19

Emma K.

Without
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04/18/19

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