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Comments on Arbeitsblatt 5 from o.Prof.Dr.F.L.Bauer.

by Edsger W.Dijkstra and C.S.Scholten

For the integer function m(x, y) of the integer arguments x and y , given by

m(x,y) = (x ≥ y → x+1 ▯ x < y → y-1)(1)
it is required to prove that
(x = y → y+1 ▯ x ≠ y → m(x, m(x-1, y+1))) = m(x, y)(2).

Proof. From (1) follows that the first alternative in the left-hand side of (2) may be replaced by x = y → m(x, y) ; the proof is complete when we show that the second alternative can be replaced by x ≠ y → m(x, y) . This letter replacement is allowed as we shall prove

m(x, m(x-1, y+1)) = m(x, y)(3)
From (1) follows
X ≥ y ⇒ (m(x, x) = m(x, y))(4)
From (1) it also follows that the only possible values of m(x-1, y+1) are x and y . If m(x-1, y+1) = y , (3) holds trivially. Otherwise we have m(x-1, y+1) ≠ y , which implies —on account of (1)—
m(x-1, y+1) = x and x-1 ≥ y+1 .(5)
As (5) implies x ≥ y , we conclude by (4) and (5) that (3) then holds as well. (End of proof.)

The above proof has been given, because 33(!) lines of formal text, as dedicated to a proof of this theorem in said Arbeitsblatt 3, is to our tastes a bit unwieldy. We appreciate that the proof in said Arbeitsblatt 3 has been conducted in terms of a limited repertoire of operations that seem suitable for mechanization. When, however, the length of such proofs and their obvious tedium are presented —as is often the case— as conclusive evidence for the necessity of such mechanizations, it should be clear that at least the above example has failed to convince us of such necessity.

Plataanstraat 5prof.dr.Edsger W.Dijkstra
NL-4565 NUENENBurroughs Research Fellow
The Netherlands

Transcribed by Martin P.M. van der Burgt
Last revision 2015-01-24 .