let f1, f2 be Function of L,(L opp); :: thesis: ( ( for x being Element of L holds f1 . x = 'not' x ) & ( for x being Element of L holds f2 . x = 'not' x ) implies f1 = f2 )

assume A2: ( ( for x being Element of L holds f1 . x = 'not' x ) & ( for x being Element of L holds f2 . x = 'not' x ) & not f1 = f2 ) ; :: thesis: contradiction

assume A2: ( ( for x being Element of L holds f1 . x = 'not' x ) & ( for x being Element of L holds f2 . x = 'not' x ) & not f1 = f2 ) ; :: thesis: contradiction

now :: thesis: for x being Element of L holds f1 . x = f2 . x

hence
contradiction
by A2, FUNCT_2:63; :: thesis: verumlet x be Element of L; :: thesis: f1 . x = f2 . x

thus f1 . x = 'not' x by A2

.= f2 . x by A2 ; :: thesis: verum

end;thus f1 . x = 'not' x by A2

.= f2 . x by A2 ; :: thesis: verum