Sindbad~EG File Manager

<p>This example is the algebraic part of Lagrange's theorem.</p>
<p><strong>Theorem: </strong><em>Let G be a group and H a subgroup of G. Let a be a member of G. Then there is a one-one correspondence between H and the coset Ha.</em></p>
<p>This file constructs a map from a subgroup H of a group G
  to the coset Ha,  and also constructs its inverse.
  The map in question takes <em>x</em> to <em>x*a</em>,  and the inverse
  takes <em>z</em> to <em>z*i(a)</em>,  where <em>i(a)</em> denotes the inverse of <em>a</em>.  
  Otter-lambda uses lambda-unification to construct these maps,
  which are given by the lambda terms lambda(x,x*a) and lambda(z,z*i(a))
respectively.  </p>
<p>Answer literals  (which use the symbol $ANS)  are
  used in Otter and Otter-lambda just to provide information to the 
  user.  By using $ANS in the input file, we cause Otter-lambda to show
  us the lambda terms it constructed in find a proof.</p>
<p>We do not use a unary predicate for G(x).   This is not an error or oversight.  For an explanation 
of why we do not need G(x),  see <a href="../ImplicitTyping"><em>Implicit Typing</em></a></p>
<p> The first axiom is always used when equality is involved:
    
1 [] x=x.</p>
<p>The next axiom is one way of formulating the axiom for group inverse, 
    without a symbol for the identity element:
    
2 [] u*x=v|u!=v*i(x).</p>
<p>The next axiom defines the existential quantifier.  It says "If the predicate
    Z holds of any object w then there exists an x such that Z(x)".   Here
"exists" is treated as a constant that applies to a predicate;  predicates
    are given by lambda-terms.   What we would usually write as 
"there exists an x such that P(x)"  is written here in the form 
    exists(lambda(x,Ap(P,x))).
    
3 [] -Ap(Z,w)|exists(lambda(x,Ap(Z,x))).</p>
<p>We now define f to be the characteristic function of the coset Ha:
      
4 [] f(w)=exists(lambda(x,and(H(x),x*a=w))).</p>
<p> Now, the theorem in question is that there exists an X such that X maps H into Ha,
      and an inverse map Y such that for a in H, we have  Y(X(a)) = a.  Using Ap
      to express function application, this would be "for all z in H, Ap(X,z) is in H and  Ap(Y,Ap(X,u)) = z". 
      As usual in a clause-based prover, we negate the goal, so the first z becomes a Skolem term g(X)
      and the second z becomes another Skolem term r(X,Y).  We could equally well have made
      both occurrences of z into r(X,Y), but we didn't do that.
      X and Y are variables, and the negated goal says that g(X) is in H, and "either Ap(X,g(X)) 
      is not in the coset H(g(X)),  or it's not the case that Ap(Y,Ap(X,r(X,Y))) = r(X,Y)."  But 
      of course we use f(z) to express that z belongs to the coset Hz.   Our negated
      goal then is expressed by the next two clauses:    </p>
<p>5 [] H(g(X)).    </p>
<p>6 [] -f(Ap(X,g(X)))|$ANS(X)|Ap(X,Ap(Y,r(X,Y)))!=r(X,Y)|$ANS(Y).</p>
<p>We put the answer literals in, as explained above, so we can see the values produced for X and Y.      </p>
<p>Otter-lambda begins the proof by starting with line 6 and using line 4 to expand the definition of f:      </p>
<p>9 [6,demod,4] -exists(lambda(x,and(H(x),x*a=Ap(y,g(y)))))|$ANS(y)|Ap(y,Ap(z,r(z)))!=r(z)|$ANS(z).</p>
<p>Then it uses the definition of "exists" from line 3.  The variable Z in line 3
          gets instantiated to lambda(x,and(H(x),x*a = Ap(X,g(X)))),  and the first 
          literal in line 3, which was -Ap(Z,w),  then beta-reduces,  now that Z is a lambda-term,
          to  -and(H(w),w*a = Ap(X,g(X))).   Otter-lambda renames the variables when producing a new 
          clause, so in the following line, what was w becomes z and what was X becomes x:</p>
<p>21 [binary,9.1,3.2,demod,beta] $ANS(x)|Ap(x,Ap(y,r(y)))!=r(y)|$ANS(y)| -and(H(z),z*a=Ap(x,g(x))).</p>
<p>Now the literal containing a negated "and" is split into two literals.  This is an inference
            rule that belongs to Otter-lambda, but not to Otter.  It is activated by the set(lambda) command.</p>
<p>22 [split -and,21] $ANS(x)|Ap(x,Ap(y,r(y)))!=r(y)|$ANS(y)| -H(z)|z*a!=Ap(x,g(x)).</p>
<p>The literal -H(z) resolves with line 5, making z = g(X).  The last literal in 22 then would 
                become g(X) *a != Ap(x,g(x)).  Otter-lambda tries to resolve this with the equality axiom 1,
                so it has to unify Ap(x,g(x)) with g(X)*a.  Lambda-unification succeeds, taking x = lambda(u,u*a)
                and X = x. With that value of x,  Ap(x,g(x)) beta-reduces to g(x)*a = g(lambda(u,u*a)) *a.   
                So the last two literals of 22 disappear and x gets the value lambda(u,u*a).  That leaves 
                only one literal in 22 besides the answer literals, and since x has been instantiated to 
                lambda(u,u*a), the left-hand side of that literal, namely Ap(x,Ap(y,r(y))),  will now 
                beta-reduce, becoming Ap(y,r(y))*a.   As usual variables are renamed in the new clause:</p>
<p>25 [binary,22.3,1.1,demod,beta,unit_del,5] $ANS(lambda(x,x*a))|Ap(y,r(y))*a!=r(y)|$ANS(y).</p>
<p>Now, that last literal is resolved with first literal of axiom 2, the axiom for group inverse.
                    Otter-lambda has to resolve Ap(y,r(y))*a = r(y) with u *x = v.  This succeeds with the 
                    assignments u:= Ap(y,r(y)), x:= a, v:=r(y).  The second literal of 2, u != v*i(x),
                    then becomes Ap(y,r(y)) != r(y)*i(a) as seen in the next line of the proof:
                    
27 [binary,25.1,2.1] $ANS(lambda(x,x*a))|$ANS(y)|Ap(y,r(y))!=r(y)*i(a).
                    
Using the equality axiom again, Otter-lambda unifies the two sides of this equality
                    to reach the final contradiction, producing the map y := lambda(z,z*i(a)) as shown 
                    in the answer literal.                  </p>
<p>28 [binary,27.1,1.1] $ANS(lambda(x,x*a))|$ANS(lambda(z,z*i(a))).
                      
Since this clause contains only answer literals, it represents a contradiction,
                    completing the proof. </p>