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 <p><strong>Proof that x &le; 0 implies x = 0 in PA, using the definition of &le; in terms of existence.</strong></p>
<p>One should read the commentary on le.in before reading this one.</p>
<p>This is another illustration of the use of &quot;exists&quot; in  lambda logic. Proceeding by contradiction as resolution provers do, we assume a &le; 0 but a &ne; 0. Then by the definition of &le;, there is some u such that a + u = 0. That u is e(lambda(x,a+x = 0)), and this point is reached in line [17] of the Otter-lambda proof below. Now in Peano Arithmetic, of course, variables range over natural numbers, so a + u = 0 is only possible if a and u are both 0. That fact, however, is not directly given in the axioms. The last step in the Otter-lambda proof below is a bit complicated. It uses <em>one</em> of the two induction axioms, but not both! Writing line [17] as a + c = 0 for a moment, using a constant c for the term e(lambda(x,a+x=0)) for readability, lambda unification is used to unify a + c = 0 with ap(y,z) from the third literal of line [6]. The masking term will be c, and y will get the value lambda(z,a + z = 0). The result of the resolution inference comes from the first two literals of [6], namely ap(y,0) | ap(y,s(g(y))); with y having the stated value, a beta-reduction is possible, and the result of the resolution inference will be a+0 = 0 | a + s(g(y)) = 0. The literal a+0 demodulates to a = 0, which contradicts [1] and is removed by unit deletion. That leaves a + s(g(y)) = 0, which demodulates to s(a + g(y)) = 0, and then contradicts [3]. Hence it too is removed by unit deletion, completing the proof without ever needing the induction hypothesis! </p>
<p>  <br>
1 [] a!=0. <br>
3 [] s(x)!=0. <br>
6 [] ap(y,0)|ap(y,s(g(y)))| -ap(y,z). <br>
9 [] -exists(lambda(x,ap(Z,x)))|ap(Z,e(Z)). <br>
11 [] x+0=x. <br>
12 [] x+s(y)=s(x+y). <br>
13 [] (u&lt;=v)=exists(lambda(x,u+x=v)). <br>
14 [] a&lt;=0. <br>
16 [14,demod,13] exists(lambda(x,a+x=0)). <br>
17 [binary,16.1,9.1,demod,beta] a+e(lambda(x,a+x=0))=0. <br>
18 [binary,17.1,6.3,demod,beta,11,beta,12,unit_del,1,3] $F.  </p>