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Proof that if f is a characteristic function, then charq f = p1

----> UNIT CONFLICT at  66.78 sec ----> 5520 [binary,5519.1,4556.1] $ans(noteq(isp1 badf,badf)).

Length of proof is 8.  Level of proof is 6.

Proof



1 [reference] k(x) y=x.
4 [reference] pair(x,y) z=pair(x z,y z).
5 [reference] pair(p1 x,p2 x)=x.
6 [reference] abst x y z=x k(z) (y z).
15 [reference] x!=y|eq pair(x,y) =p1.
16 [reference] x=y|eq pair(x,y) =p2.
17 [reference] x=y|x noteq(x,y) !=y noteq(x,y).
19 [reference] x y!=p1|isp1 x y =p1.
20 [reference] x y=p1|isp1 x y =p2.
30 [reference] charq=abst k(eq) pair(pair(p1,p2),isp1).
32 [assumption to refute] p.
33 [assumption to refute] -p|charq badf !=p1.
34 [assumption to refute] badf x=p1|badf x =p2|$ans(x).
35 [ur,33,32,demod,30,6,1,4,4,5] eq pair(badf,isp1 badf)!=p1.
56 [ur,35,15,flip.1] isp1 badf!=badf.
65 [para_into,35.1.1,16.2.1,unit_del,56] p2!=p1.
75 [para_into,65.1.2,20.1.2,flip.1] x y!=p2|isp1 x y =p2.
78 [ur,56,17] isp1 badf noteq(isp1 badf,badf)!=badf noteq(isp1 badf,badf).
4548 [para_into,78.1.1,75.2.1,flip.1] badf noteq(isp1 badf,badf)!=p2.
4556 [para_into,78.1.1,19.2.1,flip.1] badf noteq(isp1 badf,badf)!=p1.
5519 [ur,4548,34] badf noteq(isp1 badf,badf)=p1|$ans(noteq(isp1 badf,badf)).
5520 [binary,5519.1,4556.1] $ans(noteq(isp1 badf,badf)).



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