allzermalmer

Truth suffers from too much analysis

Posts Tagged ‘Conjunction’

False Hypotheses & True Predictions

Posted by allzermalmer on June 23, 2016

In logic, a conjunction is a logical connective that connects two separate propositions. For example, say we have the propositions ‘The Golden State Warriors won the Western Conference Championship of the NBA in 2016’ and ‘The Cleveland Cavaliers won the Eastern  Conference Championship of the NBA in 2016’. We can represent each of those propositions, respectively, as P and Q.

The logical connective of conjunction would combine each of these two separate propositions together. Each of these propositions would be known as a conjunct that makes up a conjunction. Conjunct of P and conjunct of Q make up the conjunction of ‘Both The Golden State Warriors won the Western Conference Championship of the NBA in 2016 & The Cleveland Cavaliers won the Eastern Conference Championship of the NBA in 2016’. This can be represented as ‘P&Q’.

A conjunction is only true when each conjunct is true. A conjunction is false when either one of the conjuncts is false or both conjuncts are false. In the example presented, it is true that both teams won the Conference championships in 2016. So the conjunction is a true proposition.

Logic tells us that from false hypotheses, or hypothesis, that true predictions follow from it.  Suppose that P means ‘The Golden State Warriors won the NBA Championship in 2015’ and that Q means ‘The Golden State Warriors won the NBA Championship in 2016’. From these two propositions, we can form the conjunction of ‘Both The Golden State Warriors won the NBA Championship in 2015 & The Golden State Warriors won the NBA Championship in 2016’. This can be represented as ‘P&Q’.

Taking ‘P&Q’ as a hypothesis, we can prove that some propositions follow from that hypothesis. One of these propositions that follow is P.  So from the hypothesis of ‘Both The Golden State Warriors won the NBA Championship in 2015 & The Golden State Warriors won the NBA Championship in 2016’ that it necessarily follows by rules of logic that ‘The Golden State Warriors won the NBA Championship in 2015’.

Suppose ‘P&Q’ then necessarily follows ‘P’.

The hypothesis presented is false, P&Q is false. One of the conjuncts is false, Q is false. One of the conjuncts is true, P is true. So the conjunction is false. But from this false hypothesis, we find that a true conclusion follows from it.

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Fundamental Tautologies

Posted by allzermalmer on September 29, 2013

First I shall list all the truth tables for basic logical operators. They shall each be given their own symbol as an operator. I will give both two different symbols for them, one for symbolic notation and one in polish notation.

Φ and Ψ will be used as meta-variables, which may be replaced by propositions at any time.

Meta-Variable for proposition Φ:
Given that Φ=True then Φ=True.
Given that Φ=False then Φ=False.

Symbolic (~) and Polish (N): Not..
Given that Φ=True then NΦ=False or (~Φ=False).
Given that Φ=False then NΦ=True or (~Φ=True).

Symbolic(&) and Polish (K): Both…and…
Given that Φ=True and Ψ=True, then KΦΨ=True or (Φ&Ψ)=True.
Given that Φ=True and Ψ=False, then KΦΨ=False or (Φ&Ψ)=False.
Given that Φ=False and Ψ=True, then KΦΨ=False or (Φ&Ψ)=False.
Given that Φ=False and Ψ=False, then KΦΨ=False or (Φ&Ψ)=False.

Symbolic (↓) and Polish (X): Neither…nor…
Given that Φ=True and Ψ=True, then XΦΨ=False or (Φ↓Ψ)=False.
Given that Φ=True and Ψ=False, then XΦΨ=False or (Φ↓Ψ)=False.
Given that Φ=False and Ψ=True, then XΦΨ=False or (Φ↓Ψ)=False.
Given that Φ=False and Ψ=False, then XΦΨ=True or (Φ↓Ψ)=True.

Symbolic (<->) and Polish (E): …if and only if…
Given that Φ=True and Ψ=True, then EΦΨ=True or (Φ<->Ψ)=True.
Given that Φ=True and Ψ=False, then EΦΨ=False or (Φ<->Ψ)=False.
Given that Φ=False and Ψ=False, then EΦΨ=False or (Φ<->Ψ)=False.
Given that Φ=False and Ψ=False, then EΦΨ=True or (Φ<->Ψ)=True.

Symbolic (v) and Polish (A): Either…or…both
Given that Φ=True and Ψ=True, then AΦΨ=True or (ΦvΨ)=True.
Given that Φ=True and Ψ=False, then AΦΨ=True or (ΦvΨ)=True.
Given that Φ=False and Ψ=True, then AΦΨ=True or (ΦvΨ)=True.
Given that Φ=False and Ψ=False, then AΦΨ=False or (ΦvΨ)=False.

Symbolic (↑) and Polish (D): Not both…and…
Given that Φ=True and Ψ=True, then DΦΨor (Φ↑Ψ)=False.
Given that Φ=True and Ψ=False, then DΦΨ or (Φ↑Ψ)=True.
Given that Φ=False and Ψ=True, then DΦΨ or (Φ↑Ψ)=True.
Given that Φ=False and Ψ=False, then DΦΨ or (Φ↑Ψ)=True.

Symbolic (->) and Polish (C): If…then…
Given that Φ=True and Ψ=True, then CΦΨ or (Φ->Ψ)=True.
Given that Φ=True and Ψ=False, then CΦΨ or (Φ->Ψ)=False.
Given that Φ=False and Ψ=True, then CΦΨ or (Φ->Ψ)=True.
Given that Φ=False and Ψ=False, then CΦΨor (Φ->Ψ)=True.

Tautologies:

Symbolic (&) and Polish (K): Both…and…
~(Φ&~Φ)=NKΦNΦ
~(~Φ&Φ)=NKNΦΦ

Symbolic (↓) and Polish (X):Neither…nor…
~(~Φ↓Φ)=NXNΦΦ
~(Φ↓~Φ)=NXΦNΦ

Symbolic (<->) and Polish (E):…if and only if…
(Φ<->Φ)=EΦΦ
(~Φ<->~Φ)=ENΦNΦ

Symbolic (v) and Polish (A):Either…or…both
(Φv~Φ)=AΦNΦ
(~ΦvΦ)=ANΦΦ

Symbolic (↑) and Polish (D):Not both…and…
(~Φ↑Φ)=DNΦΦ
(Φ↑~Φ)=DΦNΦ

Symbolic (->) and Polish (C): If…then…
(Φ->Φ)=CΦΦ
(~Φ->~Φ)=CNΦNΦ

Equivalence:

The order of these equivalence follow those above: (&), (↓), (<->), (v), (->), (↑)

(K) (Φ&Ψ): (Φ&Ψ), (~Φ&~Ψ), ~(Φ&~Ψ)&~(~Φ&Ψ), ~(~Φ&~Ψ), ~(Φ&~Ψ), ~(Φ&Ψ)

(X) (Φ↓Ψ): (~Φ↓~Ψ), (Φ↓Ψ), ~((~Φ↓~Ψ)↓(Φ↓Ψ)), ~(Φ↓Ψ), ~(~Φ↓Ψ), ~(~Φ↓~Ψ)

(A) (ΦvΨ): ~(~Φv~Ψ), ~(ΦvΨ), ~(Φv~Ψ)v~(ΦvΨ), (ΦvΨ), (~ΦvΨ), (~Φv~Ψ)

(D) (Φ↑Ψ): ~(Φ↑Ψ), ~(~Φ↑~Ψ), ~(Φ↑Ψ)↑(Ψ↑~Φ), (~Φ↑~Ψ), (Φ↑~Ψ), (Φ↑Ψ)

(C) (Φ->Ψ): ~(Φ->~Ψ), ~(~Φ->Ψ), ~((Φ->Ψ)->~(Ψ->Φ)), (~Φ->Ψ), (Φ->Ψ), (Φ->~Ψ)

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Proof of Modus Tollens

Posted by allzermalmer on July 28, 2013

Language

(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 Modus Tollens: Given the conditional claim that the consequent is true if the antecedent is true, and given that the consequent is false, we can infer that the antecedent is also false.

(If p implies q & ~q, then necessarily true that ~p)

Premise (1) p implies q (Hypothesis)
Premise (2) ~q (Hypothesis)
(3) q implies Ø ((2) and of Definition ~)
(4) p (Hypothesis)
(5) p implies q (Reiteration of (1))
(6) q (Modus Ponens by (4) and (5))
(7) q implies Ø (Reiteration of (3))
(8) Ø (Modus Ponens by (6) and (7))
(9) p implies Ø ( Conditional Proof by  (5) through (8))
Conclusion (10) ~p ((9) and Definition of ~)

Shortened version, with some steps omitted, would go as follows.

P (1) p implies q
P (2) ~q
(3) q implies Ø ((2) and Definition of ~)
(4) p (Hypothesis)
(5) q (Modus Ponens by (1) and (4))
(6) Ø (Modus Ponens by (3) and (5))
(7) p implies Ø (Conditional Proof by (3) through (6))
C (8)  ~p ((7) and Definition ~)

Here is an even shorter proof of Modus Tollens, and it only requires the rule of inference of Hypothetical Syllogism:

(1) p implies q (Hypothesis)
(2) q implies Ø (Hypothesis)
(3) p implies Ø (Hypothetical Syllogism by (1) and (2))
(4) ~p (Reiteration of (3) by Definition of ~)

So we have proved that If p implies q and ~q, then ~p is necessarily true.

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3 Value Logic

Posted by allzermalmer on May 10, 2013

I am going to use Polish Notation in expressing these truth tables of 3 value logic and 2 value logic. Lower case letters are variables: x, y, z, …
Capital Letters represent logical operators: N, A, K, E, C

Nx = ~x
Axy = x v y
Kxy = x & y
Cxy = x → y
Exy = x ↔ y

This notation is explicated in the text book Formal Logic by logican A.N. Prior in the late 1950’s to early 1960’s. The notationw as used by Polish logican Jan Lukasiewicz. Lukasiewicz was one of the first logicans to formally organize a three value logic. The logical matrix or logical matrices of both 2 value logic and 3 value logic are presented.

2 value logic uses 1 and 0.
3 value uses 1, 1/2, and 0.

1 stand for true.
1/2 stands for indeterminate.
0 stands for false.

I have put in bold those portions of truth tables in 3 value logic that do not have a similar truth table in 2 value.

Affirmation & Negation (x & Nx) : 2 value

  • (1) If x = 1 then Nx = 0.
    (2) If x = 0 then Nx = 1.

Affirmation & Negation (x & Nx) : 3 value

  • (1) If x = 1 then Nx = 0.
  • (2) If x = 1/2 then Nx = 1/2
  • (3) If x = 0 then Nx = 1

Conditional (Cxy) : 2 value

  • (1) If x = 1 and y = 1, then Cxy = 1
  • (2) If x = 1 and y = 0, then Cxy = 0
  • (3) If x = 0 and y = 1, then Cxy = 1
  • (4) If x =0 and y = 0, then Cxy = 1

Conditional (Cxy) : 3  value

  • (1) If x = 1 and y = 1, then Cxy = 1
  • (2) If x = 1 and y = 1/2, then Cxy = 1/2
  • (3) If x = 1 and y = 0, then Cxy = 0
  • (4) If x = 1/2 and y = 1, then Cxy = 1
  • (5) If x = 1/2 and y = 1/2, then Cxy = 1
  • (6) If x = 1/2 and y = 0, then Cxy = 1/2
  • (7) If x = 0 and y = 1, then Cxy = 1
  • (8) If x = 0 and y = 1/2, then Cxy = 1
  • (9) If x = 0 and y = 0, then Cxy = 1

Conjunction (Kxy) : 2 value

  • (1) x = 1 and y = 1, then Kxy = 1
  • (2) x = 1 and y = 0, then Kxy = 0
  • (3) x = 0 and y = 1, then Kxy = 0
  • (4) x = 0 and y = 0, then Kxy = 0

Conjunction (Kxy) : 3 value

  • (1) If x = 1 and y = 1, then Kxy = 1
  • (2) If x = 1 and y = 1/2, then Kxy = 1/2
  • (3) If x = 1 and y = 0, then Kxy = 0
  • (4) If x = 1/2 and y = 1, then Kxy = 1/2
  • (5) If x = 1/2 and y = 1/2, then Kxy = 1/2
  • (6) If x = 1/2 and y = 0, then Kxy = 1/2
  • (7) If x = 0 and y = 1, then Kxy = 0
  • (8) If x = 0 and y = 1/2, then Kxy = 0
  • (9) If x = 0 and y = 0, then Kxy = 0

Disjunction (Axy) : 2 value

  • (1) If x = 1 and y = 1, then Axy = 1
  • (2) If x = 1 and y = 0, then Axy = 1
  • (3) If x = 0 and y = 1, then Axy = 1
  • (4) If x = 0 and y = 0, then Axy = 0

Disjunction (Axy) : 3 value

  • (1) If x = 1 and y = 1, then Axy = 1
  • (2) If x = 1 and y = 1/2, then Axy = 1
  • (3) If x = 1 and y = 0, then Axy = 0
  • (4) If x = 1/2 and y = 1, then Axy = 1
  • (5) If x = 1/2 and y = 1/2, then Axy = 1/2
  • (6) If x = 1/2 and y = 0, then Axy = 1/2
  • (7) If x = 0 and y = 1, then Axy = 1
  • (8) If x = 0 and y = 1/2, then Axy = 1/2
  • (9) If x = 0 and y = 0, then Axy = 0

Biconditional (Exy) : 2 value

  • (1) If x = 1 and y = 1, then Exy = 1
  • (2) If x = 1 and y = 0, then Exy = 0
  • (3) If x = 0 and y = 1, then Exy = 0
  • (4) If x = 0 and y = 0, then Exy = 1

Biconditional (Exy) : 3 value

  • (1) If x = 1 and y = 1, then Exy = 1
    (2) If x = 1 and y = 1/2, then Exy = 1/2
    (3) If x = 1 and y = 0, then Exy = 0
    (4) If x = 1/2 and y = 1, then Exy = 1/2
    (5) If x = 1/2 and y = 1/2, then Exy = 1
    (6) If x = 1/2 and y = 0, then Exy = 1/2
    (7) If x = 0 and y = 1, then Exy = 0
    (8) If x = 0 and y = 1/2, then Exy = 1/2
    (9) If x = 0 and y = 0, then Exy = 1

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