Is this simplest version of the Liar Paradox: ∃LP (LP ↔ ⊢¬LP) refutable?
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peteolcott
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May 27, 2019, 11:23:13 AM5/27/19
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LHS(LP) and RHS(⊢¬LP) of this expression: (LP ↔ ⊢¬LP)
(1) If LP was true then the LHS would be true and the RHS would be false.
(2) If ¬LP was true then the LHS would be false and the RHS would be true.
Thus in neither case: LHS ↔ RHS, ∴ ¬∃LP.
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Copyright 2019 Pete Olcott All rights reserved
"Great spirits have always encountered violent
opposition from mediocre minds." Albert Einstein