Proposed Resolution of The Liar Paradox
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Here, we argue, using elementary set theory, that "This sentence is false" (the so-called Liar sentence) is neither a true nor a false sentence. It can, however, be classed as a sentence of "indeterminate truth value." Other examples of such sentences include all questions and instructions.
THEOREM
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ALL(s):ALL(t):ALL(f):ALL(m):[Set(s) & Set(t) & Set(f) & Set(m)
=> [ALL(a):[a e t => a e s] (Subsets of s)
& ALL(a):[a e f => a e s]
& ALL(a):[a e m => a e s]
& ALL(a):[a e s => [a e t | a e f | a e m] & ~[a e t & a e f] & ~[a e t & a e m] & ~[a e f & a e m]] (Trichotomy)
=> ALL(b):[b e s => [[b e t <=> b e f] => b e m]]]]
Where we can interpret:
s = a set of sentences
t = the subset of true sentences in s
f = the subset of false sentence in s
m = the set sentences "of indeterminate truth value") in s
Notation: '|' = OR-operator
Dan Christensen
2023-07-23
2023-09-17
PROOF
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Let s, t, f, m be arbitrary sets
1 Set(s) & Set(t) & Set(f) & Set(m)
Premise
2 Set(s)
Split, 1
3 Set(t)
Split, 1
4 Set(f)
Split, 1
5 Set(m)
Split, 1
Suppose t, f and m are subsets of s on which the trichotomy rules hold.
(See formal proof of the existence of such trichotomies on ANY set here.)
6 ALL(a):[a e t => a e s]
& ALL(a):[a e f => a e s]
& ALL(a):[a e m => a e s]
& ALL(a):[a e s => [a e t | a e f | a e m] & ~[a e t & a e f] & ~[a e t & a e m] & ~[a e f & a e m]]
Premise
Trichotomy rules:
7 ALL(a):[a e s => [a e t | a e f | a e m] & ~[a e t & a e f] & ~[a e t & a e m] & ~[a e f & a e m]]
Split, 6
Let x be the Liar sentence
Suppose...
8 x e s
Premise
Apply the trichotomy rule for x
9 x e s => [x e t | x e f | x e m] & ~[x e t & x e f] & ~[x e t & x e m] & ~[x e f & x e m]
U Spec, 7
10 [x e t | x e f | x e m] & ~[x e t & x e f] & ~[x e t & x e m] & ~[x e f & x e m]
Detach, 9, 8
11 x e t | x e f | x e m
Split, 10
12 ~[x e t & x e f]
Split, 10
Suppose, as is usually done, that the Liar sentence is a true sentence if and only if it is
a false sentence.
13 x e t <=> x e f
Premise
14 [x e t => x e f] & [x e f => x e t]
Iff-And, 13
15 x e t => x e f
Split, 14
16 x e f => x e t
Split, 14
Prove x is not a true sentence
Suppose to the contrary...
17 x e t
Premise
18 x e f
Detach, 15, 17
19 ~~[x e t => ~x e f]
Imply-And, 12
20 x e t => ~x e f
Rem DNeg, 19
21 ~x e f
Detach, 20, 17
Obtain the contradiction:
22 x e f & ~x e f
Join, 18, 21
As Required:
23 ~x e t
4 Conclusion, 17
Prove x is not a false sentence
Suppose to the contrary...
24 x e f
Premise
25 x e t
Detach, 16, 24
26 ~~[x e t => ~x e f]
Imply-And, 12
27 x e t => ~x e f
Rem DNeg, 26
28 ~x e f
Detach, 27, 25
Obtain the contradiction:
29 x e f & ~x e f
Join, 24, 28
As Required:
30 ~x e f
4 Conclusion, 24
Prove that x must be sentence of indeterminate truth value
Apply the trichotomy rule
31 ~[x e t | x e f] => x e m
Imply-Or, 11
32 ~~[~x e t & ~x e f] => x e m
DeMorgan, 31
33 ~x e t & ~x e f => x e m
Rem DNeg, 32
34 ~x e t & ~x e f
Join, 23, 30
x is a sentence of indeterminate truth value
35 x e m
Detach, 33, 34
As Required:
36 [x e t <=> x e f] => x e m
4 Conclusion, 13
As Required:
37 ALL(b):[b e s => [[b e t <=> b e f] => b e m]]
4 Conclusion, 8
As Required:
38 ALL(a):[a e t => a e s]
& ALL(a):[a e f => a e s]
& ALL(a):[a e m => a e s]
& ALL(a):[a e s => [a e t | a e f | a e m] & ~[a e t & a e f] & ~[a e t & a e m] & ~[a e f & a e m]]
=> ALL(b):[b e s => [[b e t <=> b e f] => b e m]]
4 Conclusion, 6
As Required:
39 ALL(s):ALL(t):ALL(f):ALL(m):[Set(s) & Set(t) & Set(f) & Set(m)
=> [ALL(a):[a e t => a e s]
& ALL(a):[a e f => a e s]
& ALL(a):[a e m => a e s]
& ALL(a):[a e s => [a e t | a e f | a e m] & ~[a e t & a e f] & ~[a e t & a e m] & ~[a e f & a e m]]
=> ALL(b):[b e s => [[b e t <=> b e f] => b e m]]]]
4 Conclusion, 1