print(`Let A, B be forward and M be backward shift`):
print(`operators on a, b and m respectively.`):

print(`Denote the identity on the right-hand-side of Thm 10' by R`):

print(`Then we verify that R(a,b,m) satisfies the recurrence`):
print(``):

print((X(m-1,a,b)*X(m-1,a+1,b+1)-X(m-1,a+1,b)*X(m-1,a,b+1))/X(m,a,b)/X(m-2,a+1,b+1)=1,`or`):

print(M(R)*MAB(R)/R/MMAB(R)-AM(R)*BM(R)/R/MMAB(R)=1):
print(`Note that:`):

a1:=(3*b+2*m-1)!*(2*a+2*m-b)!*(2*b-a+2*m)!*(3*a+2*m-1)!/m!/(3*b+3*m-1)!/(a+b+m-1)!/(3*a+3*m-1)!:
print(`M(R)/R`,`=`,a1):

a2:=product(i!*(3*b+3*i-1)!/(3*b+2*i-1)!,i=0..m-1)*
    product((a+b+i-1)!*(3*a+3*i-1)!/(2*a+2*i-b)!,i=1..m)*
    product(1/(2*b-a+2*i-1)!/(3*a+2*i+2)!,i=0..m-1):
print(MA(R),`=`,a2):
print(`and`):

a3:=(3*b+2*m-1)!*(2*a-b)!*(2*b-a+2*m)!*(3*a+2*m-1)!/m!/(3*b+3*m-1)!/(a+b-1)!/(3*a-1)!*product((3*a+2*i-1)!/(3*a+2*i+2)!,i=0..m-1)*
    product((2*b-a+2*i)!/(2*b-a+2*i-1)!,i=0..m-1):
:
print(MA(R)/R,`=`,a3):

Print(`Therefore,`):
b1:=simplify(a1/subs(a=a+1,b=b+1,m=m-1,a1)):
print(M(R)/R/ABM(MR/R),`=`,b1):
print(`and`):
a4:=(3*b+2*m-1)!*(2*a-b)!*(2*b-a+2*m)!*(3*a+2*m-1)!/m!/(3*b+3*m-1)!/(a+b-1)!/(3*a-1)!:
b2:=simplify(a4/subs(b=b+1,m=m-1,a4)*(3*a+2*m-3)!*(2*b-a)!/(3*a+2*m)!/(2*b-a-1)!):
print(MA(R)/R/BM(MA(R)/R),`=`,b2):
print(`Finally,`):

print(M(R)/R/ABM(MR/R)-MA(R)/R/BM(MA(R)/R),`=`,simplify(b1-b2)):