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Is this simplest version of the Liar Paradox: ∃LP (LP ↔ ⊢¬LP) refutable?
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peteolcott
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.
--
Copyright 2019 Pete Olcott All rights reserved
"Great spirits have always encountered violent
opposition from mediocre minds." Albert Einstein
(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.
--
Copyright 2019 Pete Olcott All rights reserved
"Great spirits have always encountered violent
opposition from mediocre minds." Albert Einstein
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