∃G ∈ F (G ↔ (F ⊬ G))
Posted: Sat Apr 22, 2023 2:59 pm
∃G ∈ F (G ↔ (F ⊬ G))
There exists a G in F such that G is logically equivalent to its own unprovability in F
Logical equality
p---q---p ↔ q
T---T------T // G is true if and only if G is Unprovable.
T---F------F //
F---T------F //
F---F------T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_tab ... l_equality
Row(1) There exists a G in F such that G is true if and only if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false and there
is no such G in F.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.
When we take the informal version of this where
G asserts its own unprovability in F
The proof of G in F requires a sequence of inference
steps in F that prove that they themselves do not exist.
There exists a G in F such that G is logically equivalent to its own unprovability in F
Logical equality
p---q---p ↔ q
T---T------T // G is true if and only if G is Unprovable.
T---F------F //
F---T------F //
F---F------T // G is false if and only if G is Provable.
https://en.wikipedia.org/wiki/Truth_tab ... l_equality
Row(1) There exists a G in F such that G is true if and only if G is
unprovable in F making G unsatisfied thus untrue in F.
Row(4) There exists a G in F such that G is false if and only if G is
provable in F making G satisfied thus true in F.
If either Row(1) or Row(4) are unsatisfied then ↔ is false and there
is no such G in F.
Because the RHS of ↔ contradicts the LHS there is no such G in F.
Thus the above G simply does not exist in F.
When we take the informal version of this where
G asserts its own unprovability in F
The proof of G in F requires a sequence of inference
steps in F that prove that they themselves do not exist.