THE LIAR PARADOX: A PROPOSED RESOLUTION
***************************************
Here, we formally prove:
ALL(a):[a e sentences => [[a e true
<=> a e false] <=> a e meaningless]]
Where: sentences, true, false and meaningless are sets defined
as below.
Dan Christensen
March 24, 2021
(This formal proof was written and verified with the aid of the
author's DC Proof 2.0 software available at http://www.dcproof.com )
AXIOMS
******
Define 3 disjoint subsets of a collection of sentences: true,
false and meaningless.
true is a subset of sentences
1 ALL(a):[a e true => a e sentences]
Axiom
false is a subset of sentences
2 ALL(a):[a e false => a e sentences]
Axiom
meaningless is a subset of senstences
3 ALL(a):[a e meaningless => a e sentences]
Axiom
Every sentence is an element of one and only one of: true,
false, meaningless
4 ALL(a):[a e sentences => [a e true | a e false | a e meaningless]
& ~[a e true & a e false]
& ~[a e true & a e meaningless]
& ~[a e false & a e meaningless]]
Axiom
PROOF
*****
Let x be a sentence
5 x e sentences
Premise
Apply definitions of the
categories: true, false and meaningless
6 x e sentences => [x e true | x e false | x e meaningless]
& ~[x e true & x e false]
& ~[x e true & x e meaningless]
& ~[x e false & x e meaningless]
U Spec, 4
7 [x e true | x e false | x e meaningless]
& ~[x e true & x e false]
& ~[x e true & x e meaningless]
& ~[x e false & x e meaningless]
Detach, 6, 5
8 x e true | x e false | x e meaningless
Split, 7
9 ~[x e true & x e false]
Split, 7
10 ~[x e true & x e meaningless]
Split, 7
11 ~[x e false & x e meaningless]
Split, 7
'=>'
Prove: [x e true <=> x e false] =>
x e meaningless
Suppose sentence x
is true if and only if
it is false, as in
The Liar Paradox
12 x e true <=> x e false
Premise
13 [x e true => x e false] & [x e false => x e true]
Iff-And, 12
14 x e true => x e false
Split, 13
15 x e false => x e true
Split, 13
Prove: x is
not true
Suppose to the
contrary...
16 x e true
Premise
17 x e false
Detach, 14, 16
18 x e true & x e false
Join, 16, 17
19 x e true & x e false & ~[x e true & x e false]
Join, 18, 9
As Required:
20 ~x e true
4 Conclusion, 16
Prove: x is not
false (by contradiction)
21 x e false
Premise
22 x e true
Detach, 15, 21
23 x e true & ~x e true
Join, 22, 20
24 ~x e false
4 Conclusion, 21
Prove: x is
meaningless
25 ~[~x e true & ~x e false] | x e meaningless
DeMorgan, 8
26 ~~[~x e true & ~x e false] => x e meaningless
Imply-Or, 25
27 ~x e true & ~x e false => x e meaningless
Rem DNeg, 26
28 ~x e true & ~x e false
Join, 20, 24
29 x e meaningless
Detach, 27, 28
As Required:
30 [x e true <=> x e false] => x e meaningless
4 Conclusion, 12
'<='
Prove: x e meaningless => [x e true
<=> x e false]
Suppose...
31 x e meaningless
Premise
Prove: ~x e true
Suppose the
contrary...
32 x e true
Premise
33 x e true & x e meaningless
Join, 32, 31
34 x e true & x e meaningless
& ~[x e true & x e meaningless]
Join, 33, 10
As Required:
35 ~x e true
4 Conclusion, 32
Prove: x e true => x e false
Suppose...
36 x e true
Premise
37 ~x e true => x e false
Arb Cons, 36
38 x e false
Detach, 37, 35
As Required:
39 x e true => x e false
4 Conclusion, 36
Prove: ~x e false
Suppose to the
contrary...
40 x e false
Premise
41 x e false & x e meaningless
Join, 40, 31
42 x e false & x e meaningless
& ~[x e false & x e meaningless]
Join, 41, 11
As Required:
43 ~x e false
4 Conclusion, 40
Prove: x e meaningless => [x e true
<=> x e false]
Suppose...
44 x e false
Premise
45 ~x e false => x e true
Arb Cons, 44
46 x e true
Detach, 45, 43
As Required:
47 x e false => x e true
4 Conclusion, 44
48 [x e true => x e false] & [x e false => x e true]
Join, 39, 47
49 x e true <=> x e false
Iff-And, 48
As Required:
50 x e meaningless => [x e true <=> x e false]
4 Conclusion, 31
51 [[x e true <=> x e false] => x e meaningless]
& [x e meaningless => [x e true <=> x e false]]
Join, 30, 50
52 x e true <=> x e false <=> x e meaningless
Iff-And, 51
As Required:
53 ALL(a):[a e sentences
=> [a e true <=> a e false <=> a e meaningless]]
4 Conclusion, 5