Sindbad~EG File Manager

 <p><strong>Proof that a &le; s(0) implies a = 0 | a = s(0) in PA, using the definition of &le; in terms of existence.</strong></p>
<p>One should read the commentaries on le.in and le2.in before reading this one.</p>
<p>This is another illustration of the use of &quot;exists&quot; in  lambda logic. It is similar to le2.in but more difficult, and the Otter-lambda proof is hard to understand. An informal proof might proceed as follows: since a &le; s(0) we have a + u = s(0) for some u. Prove by induction on u that a + u &ne; s(0) unless a = 0 or a = s(0). The basis case u =0 demodulates to a &ne; s(0), which we are given. The induction step is a + s(u) &ne; s(0). Using the definition of addition this becomes s(a+u) &ne; s(0). Since successor is one-to-one, that becomes a+u &ne; 0. That can be proved as in le2.in, since we are also given a &ne; 0.</p>
<p>The Otter-lambda proof given below closely follows the beginning of this outline. At line [18] it begins the inductive proof; for readability the Skolem term g(lambda(x,a+x=s(0))) and its alphabetic variants have been replaced by a constant c for readability, but the occurence of this term indicates that the induction hypothesis is as sketched above. At line [32] then, Otter-lambda has arrived at a+c = 0, which corresponds to a+u = 0 in the above proof. Note that the definition of addition [13] was used already to derive [19]. So at line [32], it only remains to derive a contradiction from a+c = 0 and a &ne; 0. But now, unlike in le2.in, c is a Skolem term for induction, and not a Hilbert epsilon-symbol term. So it is NOT the first choice of a masking term when Otter-lambda attempts to unify [32] with the third literal in [7]. This file was created an run before the option existed to allow multiple unifiers; since the input file doesn't contain assign(max_unifiers,2), the default is 1 and no other masking term is tried. Hence the proof does NOT complete as in le2.in, the way the proof sketch above suggests it might. Instead it has a lot of strange steps, which I have verified step by step, but of which I have no intuitive understanding. Finally at the last step, it <em>does</em> complete as in le2.in, with an application of one but not both of the induction axioms and a use of the definition of addition. </p>
<p>I did run this file again with assign(max_unifiers,2) and again with assign(max_unifiers,8), but it finds this same proof. Without the line in the input file commented about nested g-terms, it does produce a similar proof, but involving nested Skolem terms and paramodulation into those terms, which is even uglier than this proof. </p>
<p>
1 [] a!=0.<br>
2 [] a!=s(0).<br>
3 [] x=x.<br>
4 [] s(x)!=0.<br>
5 [] s(x)!=s(y)|x=y.<br>
6 [] ap(y,0)| -ap(y,g(y))| -ap(y,z).<br>
7 [] ap(y,0)|ap(y,s(g(y)))| -ap(y,z).<br>
8 [] x+0=x.<br>
9 [] -ap(Z,w)|exists(lambda(x,ap(Z,x))).<br>
10 [] -exists(lambda(x,ap(Z,x)))|ap(Z,e(Z)).<br>
12 [] x+0=x.<br>
13 [] x+s(y)=s(x+y).<br>
14 [] (u&lt;=v)=exists(lambda(x,u+x=v)).<br>
15 [] a&lt;=s(0).<br>
18 [binary,15.1,7.3,demod,14,beta,14,14,beta,factor_simp] exists(lambda(x,a+x=s(0))).<br>
19 [binary,18.1,10.1,demod,beta] a+e(lambda(x,a+x=s(0)))=s(0).<br>
21 [binary,19.1,7.3,demod,beta,12,beta,13,unit_del,2] s(a+g(lambda(x,a+x=s(0))))=s(0).<br>
22 [binary,19.1,6.3,demod,beta,12,beta,unit_del,2] a+g(lambda(x,a+x=s(0)))!=s(0).<br>
32 [binary,21.1,5.1] a+g(lambda(x,a+x=s(0)))=0.<br>
34 [para_from,21.1.1,5.1.2] s(x)!=s(0)|x=a+g(lambda(y,a+y=s(0))).<br>
44 [binary,32.1,9.1,demod,beta] exists(lambda(x,a+g(lambda(y,a+y=s(0)))=0)).<br>
45 [para_from,32.1.1,22.1.1] 0!=s(0).<br>
46 [para_from,32.1.2,8.1.1.2] x+a+g(lambda(y,a+y=s(0)))=x.<br>
49 [para_from,32.1.2,1.1.2] a!=a+g(lambda(x,a+x=s(0))).<br>
90 [para_into,44.1.1.2.1,32.1.1] exists(lambda(x,0=0)).<br>
92 [para_into,90.1.1.2.2,32.1.2] exists(lambda(x,0=a+g(lambda(y,a+y=s(0))))).<br>
130 [binary,92.1,10.1,demod,beta] 0=a+g(lambda(x,a+x=s(0))).<br>
176 [para_from,34.2.2,130.1.2] 0=x|s(x)!=s(0).<br>
182 [para_from,34.2.2,49.1.2] a!=x|s(x)!=s(0).<br>
195 [binary,176.2,7.2,demod,beta,beta,unit_del,45] 0=g(lambda(x,x=s(0)))|y!=s(0).<br>
248 [binary,195.2,46.1] 0=g(lambda(x,x=s(0))).<br>
252 [para_into,248.1.1,130.1.1] a+g(lambda(x,a+x=s(0)))=g(lambda(y,y=s(0))).<br>
267 [para_from,248.1.1,4.1.2] s(x)!=g(lambda(y,y=s(0))).<br>
358 [binary,182.1,7.1,demod,beta,beta,unit_del,3] a=s(g(lambda(x,a=x)))|a!=y.<br>
370 [binary,358.2,3.1] a=s(g(lambda(x,a=x))).<br>
375 [para_from,370.1.1,252.1.1.1] s(g(lambda(x,a=x)))+g(lambda(y,a+y=s(0)))=g(lambda(z,z=s(0))).<br>
441 [binary,375.1,7.3,demod,beta,12,beta,13,unit_del,267,267] $F.</p>