Aus III:

(b + 1) + a
I
=
b + (1 + a)
II
=
b + (a + 1)
D
=
(b + a) + 1

(b + 1) + 1
D
=
b + (1 + 1)
(b + 1) + (a + 1)
D
=
((b + 1) + a) + 1
b + (1 + (a + 1))
D
=
b + ((1 + a) + 1)
D
=
(b + (1 + a)) + 1
         
}

    
(Ƒ)
d.h., hier gibt es einen arithmetischen periodischen Beweis.

b + (1 + 1) = b + (1 + 1)
b + (1 + (a + 1))
D
=
b + ((1 + a) + 1)
D
=
(b + (1 + a)) + 1
b + ((a + 1) + 1)
D
=
(b + (b + (a + 1)) + 1
         
}

    
(Ƒ)