- #36
DanTeplitskiy
- 70
- 0
Dear Rubi,
Yes, in predicate logic Argument 1 is a theorem. I know it ))
My point is that assumption R ∈ R contradicts the definition of R because if R ∈ R, R includes a member that is included in itself (R itself is such a member).
Or, symbolically: R ∈ R → ∃ y: y ∈ R ∧ y ∈ y → R ≠ {x: x∉x}: indeed, such y exists as we can take y = R
That is the second premise contradicts the first one. "Contradictory premises" logical error.
Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).
The reasoning on `legless Dan' contains the same logical error as the Russell's paradox does: the second premise contradicts the definition which is the first premise though to make the conclusion both premises are used.
Yours,
Dan
Yes, in predicate logic Argument 1 is a theorem. I know it ))
My point is that assumption R ∈ R contradicts the definition of R because if R ∈ R, R includes a member that is included in itself (R itself is such a member).
Or, symbolically: R ∈ R → ∃ y: y ∈ R ∧ y ∈ y → R ≠ {x: x∉x}: indeed, such y exists as we can take y = R
That is the second premise contradicts the first one. "Contradictory premises" logical error.
Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).
The reasoning on `legless Dan' contains the same logical error as the Russell's paradox does: the second premise contradicts the definition which is the first premise though to make the conclusion both premises are used.
Yours,
Dan