This sentence is false!

A Proposed Resolution of the Liar Paradox

Introduction

In natural language, some sentences are true. Some are false. Many sentences, like questions and instructions, are of indeterminate truth value, e.g. “What time is it?” and “Wash your hands.” As usually assumed, "This sentence is false" (the so-called Liar) is a true sentence if and only if it is a false sentence. This eliminates the possibilities that it is either a true sentence or a false sentence. Like so many others, it must be one of indeterminate truth value. Here, we will formally prove this using only basic set theory and ordinary logic.

The key concept here is that of a trichotomy defined as follows:

What is a Trichotomy?

• Division into three parts, classes, categories, etc.
  
• An instance of such a division, as in thought, structure, or object

https://www.dictionary.com/browse/trichotomy

Here, we will classify sentences as one of either of the following:

1. True sentence
2. False sentence
3. Sentence of indeterminate truth value

Outline of Proof

Here, we make use of only basic set theory and the ordinary rules of logic.

We begin the proof by considering 4 arbitrary sets:

1. s = a set of sentences (not necessarily every sentence)
2. t = the subset of true sentences
3. f = the subset of false sentences
4. m = the subset of those sentences of indeterminate truth value

Then we introduce the rules of the required trichotomy:

ALL(a):[a e s => [a e t | a e f | a e m]     (‘|’ = OR-operator)

         & ~[a e t & a e f]

         & ~[a e t & a e m]

    & ~[a e f & a e m]]

We then assume that sentence x is a true sentence if and only if it is a false sentence:

x e t <=> x e fwhere x represents the sentence “This sentence is false.” (The Liar)

Then we prove that sentence x is a sentence of indeterminate truth value:

x e m

In the final conclusion, we obtain the universal generalization:

ALL(s):ALL(t):ALL(f):ALL(m):[Set(s) & Set(t) & Set(f) & Set(m)

=> [ALL(a):[a ε t => a ε s]                   (3 Subsets of s)

& ALL(a):[a ε f => a ε s]

& ALL(a):[a ε m => a ε s]

& ALL(a):[a ε s => [a ε  t | a ε f | a ε m]         (Trichotomy Rules)

& ~[a ε t & a ε f]

& ~[a ε t & a ε m]

& ~[a ε f & a ε m]]  

=> ALL(b):[b ε s => [[b ε t <=> b ε f] => b ε m]]]]

Full text of proof (44 lines in DC Proof format)

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