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<p><strong>Proof in Peano Arithmetic that n &lt; a<sup>n </sup> &nbsp; provided 1 &lt; a </strong></p>
<p> We recommend that the reader first try to carry out a proof in PA with pencil and paper. </p>
<p>Here is the proof that Otter-lambda finds, with the Skolem term g(lambda(x,x&lt;a^x)) replaced by a constant c:</p>
<p> <br>
  2 [] x*0=0. <br>
  5 [] x^s(y)=x*x^y. <br>
  9 [] -ap(y,0)|ap(y,g(y))|ap(y,z). <br>
  10 [] -ap(y,0)| -ap(y,s(g(y)))|ap(y,z). <br>
  13 [] -(u&lt;v)|x*u&lt;x*v| -(0&lt;x). <br>
  15 [] 0&lt;s(0). <br>
  17 [] 0&lt;a. <br>
  19 [] x&lt;y|y&lt;=x. <br>
  20 [] -(y&lt;=x)| -(x&lt;y). <br>
  21 [] -(u&lt;v)| -(v&lt;=w)|u&lt;w. <br>
  22 [] -(s(0)&lt;z)| -(0&lt;y)|s(y)&lt;=z*y. <br>
  24 [] -(x&lt;=0)|x=0. <br>
  25 [] s(0)&lt;a. <br>
  26 [] x*s(y)=x*y+x. <br>
  28 [] 0+x=x. <br>
  29 [] x^0=s(0). <br>
  30 [] -(n&lt;a^n). <br>
  32 [binary,30.1,10.3,demod,beta,29,beta,5,unit_del,15] -(s( c)&lt;a*a^ c). <br>
  33 [binary,30.1,9.3,demod,beta,29,beta,unit_del,15]  c&lt;a^ c. <br>
  35 [hyper,33,13,17] a* c&lt;a*a^ c. <br>
  37 [binary,32.1,19.1] a*a^ c&lt;=s( c). <br>
  41 [hyper,37,21,35] a* c&lt;s( c). <br>
  42 [binary,41.1,20.2] -(s( c)&lt;=a* c). <br>
  45 [binary,42.1,22.3,unit_del,25] -(0&lt; c). <br>
  46 [binary,45.1,19.1]  c&lt;=0. <br>
  52 [binary,46.1,24.1]  c=0. <br>
  67 [para_from,52.1.1,32.1.2.2.2,demod,29,26,2,28] -(s( c)&lt;a). <br>
  76 [para_from,52.1.2,25.1.1.1] s( c)&lt;a. <br>
  77 [binary,76.1,67.1] $F. </strong> </p>
<p>The proof begins at lines 32 and 33, where the correct induction step and induction hypothesis are found by lambda unification. Note that the successor in the exponent has already been simplified by the appropriate demodulator. In axiom 13 we provided the information both sides of an inequality can be multiplied by the same positive quantity. With seeming foresight that is applied to produce [35], which with the (negated) induction step yields [41], a*c &lt; s(c), which in [42] is recognized as the negation s(c) &lt;= a*c. Now since 1 &lt; a is given, we are down to proving a fundamental property of the integers that has nothing to do with exponentiation. That property has been supplied, in essence, in axiom [22], so the proof now finishes quickly.</p>
<p>One might hope that it might be able to prove this theorem without axiom [22]; but if we comment out axiom [22] in the input file, the proof search terminates with sos empty. </p>