Truth suffers from too much analysis

Proof of Disjunctive Syllogism

Posted by allzermalmer on July 28, 2013


(I) Symbols: Ø = contradiction, → = conditional, and [] = Modal Operator
(II) Variables: p, q, r, p’, q’, r’. (Variables lower case)

Well Formed Formula for Language

(i) Ø and any variable is a modal sentence.
(ii) If A is a modal sentence, then []A is a modal sentence.
(iii) If A is a modal sentence and B is a modal sentence, then A implies B (A→B) is a modal sentence.

* A, B, and C are modal sentences, i.e. upper case letters are modal sentences. These upper case letters are “variables as well”. They represent the lower case variables in conjunction with contradiction, conditional, or modal operator.

So A may possibly stand for p, or q, or r. It may also possibly stand for a compound of variables and symbols. So A may stand for q, or A may stand for p→Ø, and etc.

Negation (~) = A→Ø
Conjunction (&) = ~(A→B)
Disjunction (v) = ~A→B
Biconditional (↔) = (A→B) & (B→A)

Because Ø indicates contradiction, Ø is always false. But by the truth table of material implication, A → Ø is true if and only if either A is false or Ø is true. But Ø can’t be true. So A → Ø is true if and only if A is false.

This symbol ∞ will stand for something being proved.

(1) Hypothesis (HY) : A new hypothesis may be added to a proof anytime, but the hypothesis begins a new sub-proof.

(2) Modus Ponens (MP) : If A implies B and A, then B must lie in exactly the same sub-proof.

(3) Conditional Proof (CP): When proof of B is derived from the hypothesis A, it follows that A implies B, where A implies B lies outside hypothesis A.

(4) Double Negation (DN): Removal of double negation ~~A & A lie in the same same sub-proof.

(5) Reiteration (R): Sentence A may be copied into a new sub-proof.

Proof of Disjunctive Syllogism: Because at least one disjunct must be true, by knowing one is false we can infer tat the other is true.

If either p or q and not p, then necessarily true q.

Premise (1) p v q (Hypothesis)
Premise (2) ~p (Hypothesis)
(3) ~p implies q ((1) and Definition v)
Conclusion (4) q (Modus Ponens by (2) and (3))


4 Responses to “Proof of Disjunctive Syllogism”

  1. Matt said

    Nice post! One question though: isn’t the definition of conjunction `~(A -> ~B)`?

  2. Matt said

    (The system ate my reply twice already. Let me try one more time.)
    Maybe I’m missing something, but the post says `Conjunction (&) = ~(A→B)`. Shouldn’t that be `Conjunction (&) = ~(A→~B)`? Maybe I’m just missing something — is the post correct? Thanks!

    • The post on my end lists the definition of a conjunction by the conditional shows this, which was posted in the reply you are making to:
      A&B= ~(A–>~B).

      Negation outside the brackets, with A connected to a Negated B by Conditional inside the brackets.

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