let f, g be Function of [:L,L:],L; ( ( for x, y being Element of L holds f . x,y = x "/\" y ) & ( for x, y being Element of L holds g . x,y = x "/\" y ) implies f = g )
assume that
A3:
for x, y being Element of L holds f . x,y = x "/\" y
and
A4:
for x, y being Element of L holds g . x,y = x "/\" y
; f = g
hence
f = g
by YELLOW_3:13; verum