Corrected, whoops ⊢S := S=false ⊢lnc := For all a, ¬(a^¬a) ⊢S=((S=false)=false) ⊢S=(S=true) ⊢(S=true)=(S=false) ⊢ (S=true)=/=(S=false) ⊢S =/= S ⊢S= ¬S Case 1: ⊢True=false ⊢True^true ⊢True^false ⊢True^ ¬true ⊢¬lnc Case 2: ⊢false=true ⊢True^ true ⊢True^false ⊢True^ ¬true ⊢¬lnc
On Sunday, July 12, 2020 at 9:22:29 AM UTC-6 Joseph V wrote: > Disproof of LNC, feel free to criticize: > ⊢S := S=false > ⊢lnc := For all a, ¬(a^¬a) > ⊢(S=false)=false > ⊢S=true > ⊢S=(true=false) > ⊢S=false > ⊢S^¬S > ⊢¬lnc -- You received this message because you are subscribed to the Google Groups "Metamath" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/97d07345-df2c-4706-b68c-654307fa86ben%40googlegroups.com.
