# Is this enough to conclude that G is false?

Let us assume a logical system with only 3 axioms (laws of thought):

``````Axiom 1: A statement that is true will remain true till a change is made in the system.
Axiom 2: A statement may not be true and not true.
Axiom 3: A statement must either be true or not true.
``````

Basic rules are that statements refer to the truth of other statements and can use boolean operators AND, OR and NOT to tell if-then statements.

Then, we state a statement, G:

``````G can not be proven as true.
``````

Now, let us fill in this table (with ticks and crosses):

``````G is       Provable true     Provable false     Not provable true/false
-----------------------------------------------------------------------
True            ?1                 ?2                  ?3
False           ?4                 ?5                  ?6
``````

Obviously, blank 2 and 4 cannot be, and can be crossed.

If G is true, then it states that it can not be proven true. Therefore, blank 1 is crossed out.

If G is false, then it can be proven, either as true or as false. Under our current assumption (of it being false), we can only prove it as being false.

Is this enough to conclude that G is false?

• what about slot 3? Slot 6 would be viable except that you've assumed it away in the second to last sentence. – Dave Mar 12 '15 at 14:40
• You have to take care in "handling" both provable and true in the same context ... See Gödel's Incompleteness Theorems and Tarski's undefinability theorem. – Mauro ALLEGRANZA Mar 12 '15 at 15:33
• Is statement G self-referential? If so, you need to give us rules for what happens (does the internal G evaluate first). Also in undefined cases, does the logic default to true or false? (e.g., the material conditional defaults to True in sentential logic in order to handle untested conditionals). – virmaior Mar 13 '15 at 0:45