How to interpret conflicting formal proofs about "a mod 0 = ? " - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T15:05:02Zhttp://mathoverflow.net/feeds/question/82181http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/82181/how-to-interpret-conflicting-formal-proofs-about-a-mod-0How to interpret conflicting formal proofs about "a mod 0 = ? "joro2011-11-29T14:28:06Z2011-11-29T14:51:49Z
<p>The proof assistants Coq and Isabelle give conflicting formal proofs about $a \mod 0 \qquad \forall a \in \mathbb{Z}$.</p>
<p>According to Coq
$$ a \mod 0 = 0$$
and Isabelle proves
$$ a \mod = a$$</p>
<p><code>mod</code> is the function, not a congruence.</p>
<blockquote>
<p>Which way is it?</p>
</blockquote>
<p>All the computer algebra systems I tried give an error in this case.</p>
<p>Can one derive a counter intuitive statement from the above results?</p>
<p>Both agree that integer division by $0$ is $0$ forall $\mathbb{Z}$.</p>
<p>Coq proof:</p>
<pre><code>Require Import ZArith.
Require Import Coq.ZArith.Znumtheory.
Open Scope Z_scope.
Lemma mod0: forall n:Z, n mod 0 = 0.
apply Zmod_0_r.
Qed.
</code></pre>
<p>Isabelle proof:</p>
<pre><code>theory mod0
imports Main
begin
lemma mod0: " \<forall> n \<in> \<int>. n mod (0::int) = n"
by auto
</code></pre>
http://mathoverflow.net/questions/82181/how-to-interpret-conflicting-formal-proofs-about-a-mod-0/82182#82182Answer by Andreas Blass for How to interpret conflicting formal proofs about "a mod 0 = ? "Andreas Blass2011-11-29T14:51:49Z2011-11-29T14:51:49Z<p>If $a$ mod 0 is to be defined at all (and I'm not entirely convinced that it should be), then it ought to differ from $a$ by a multiple of 0, which means to me that it ought to be $a$. But it's asserted in the question that the computer systems have a strange notion of division by 0, so they might think that everything is a multiple of 0. In this alternative "reality", everything is congruent to everything else modulo 0; so if you define $a$ mod $b$ as the smallest non-negative integer congruent to $a$ modulo $b$, then $a$ mod 0 would be 0. Personally, I refuse to buy into this alternative reality; congruence modulo 0 should mean equality. (Fortunately, I rarely use computer algebra systems, and I have never yet asked one about divisibility by 0.)</p>