Disproof of LEM
Case 1: LEM is true
?S := S=false
?S= ¬S
?True=false
?¬LEM
Case 2: ¬LEM
Let m be a third truth value
?S := S=false
?S= ¬S
?(S=True) or (s=m)
?¬(S=True)
?s=m
Case 1: LEM is true
?S := S=false
?S= ¬S
?True=false
?¬LEM
Case 2: ¬LEM
Let m be a third truth value
?S := S=false
?S= ¬S
?(S=True) or (s=m)
?¬(S=True)
?s=m
