“…For example, to show (27), it suffices to check that πΌ = β πΊ 0+ (π₯, π‘ β π ) + πΊ 0+ (π₯, π‘ β π ) 0 π (0 + , π )+πΊ β+ (π₯, π‘βπ ) 0 π (0 β , π )+πΊ + 0π (π₯, π‘βπ ) 0 π (0, π )andπΌ πΌ = β πΊ ++ (π₯ β 1, π‘ β π ) + πΊ ++ (π₯ + 1, π‘ β π ) , π ) + πΊ 0+ (π₯ β 1, π‘ β π ) + πΊ 0+ (π₯ + 1, π‘ β π ) 0 π (1 β , π ) + πΊ + 1π (π₯, π‘ β π ) 0 π ) vanish.For brevity, we shall only show that πΌ is zero. First, note that by(23), we haveπΊ 0+ (π₯, π‘ β π ) + πΊ 0+ (π₯, π‘ β π ) 2π (π₯ + 2π, π‘ β π ) + πΊ 1,2π+1 (π₯ + 2π, π‘ β π ) πΊ β+ (π₯, π‘ β π ) 0 π (0 + , π ).Note also that πΊ+ 0π = πΊ β+ by definition. From these, it follows that πΌ = βπΊ β+ (π₯, π‘ β π ) 0 π (0, π ) + πΊ + 0π (π₯, π‘ β π ) 0 π (0, π ) = 0.…”