% Double-negation translations and their proofs

% 1) (P | ~P)* = ~~(P* | ~(P*))
%              = ~~(~~P | ~(~~P))
% However, we can prove
%  ~~(A | ~A) 
% for an arbitrary atomic proposition A, not just
% a double-negated atom.  This is easier to read,
% and covers the case for a double-negated atom
% by substituting ~P for A.

proof lem : ~~(A | ~A) = begin
[ ~(A | ~A);
  [ A;
    A | ~A;
    F ];
  ~A;
  A | ~A;
  F ];
~~(A | ~ A)
end;

% 2) (((P => Q) => P) => P)* = ((P* => Q*) => P*) => P*
%                            = ((~~P => Q*) => ~~P) => ~~P
%                            = ((~~P => ~~Q) => ~~P) => ~~P
% Again, we'll leave Q arbitrary:

proof peircelaw :((~~P => Q) => ~~P) => ~~P = begin
[ ((~~P => Q) => ~~P);
  [~P;
   [~~P;
    F;
    Q];
   ~~P => Q;
   ~~P;
   F];
   ~~P];
((~~P => Q) => ~~P) => ~~P;
end;

% 3) (~(P & Q) => (~P | ~Q))* = (~(P* & Q*) => ~~(~(P*) | ~(Q*)))
%                             = (~(~~P & ~~Q) => ~~(~(~~P) | ~(~~Q)))
% But again we can prove the first line for arbitrary P and Q,
% which gives the result for double-negated atoms by substitution.

proof demorgan4 : ~(A & B) => ~~(~A | ~B) = 
begin
[~(A & B);
 [~(~A | ~B);
  [A;
   [B;
    A & B;
    F];
   ~B;
   ~A | ~B;
   F];
  ~A;
  ~A | ~B;
  F];
 ~~(~A | ~B)];
~(A & B) => ~~(~A | ~B);
end;

% 4) ((P => Q) => (~P | Q))* = ((P* => Q*) => ~~(~(P*) | (Q*)))
%                            = ((~~P => ~~Q) => ~~(~(~~P) | (~~Q)))

proof io : ((P => Q) => ~~(~P | Q)) = 
begin
[(P => Q);
 [~(~P | Q);
  [P; 
   Q;
   ~P | Q;
   F];
  ~P;
  ~P | Q;
  F];
 ~~(~P | Q)];
((P => Q) => ~~(~P | Q));
end;

% 5) ( (P => (Q | R)) => ((P => Q) | (P => R)) )* 
%  = (P* => ~~(Q* | R*)) => ~~((P* => Q*) | (P* => R*))
%  = (~~P => ~~(~~Q | ~~R)) => ~~((~~P => ~~Q) | (~~P => ~~R))

proof dist : (P => ~~(Q | R)) => ~~((P => Q) | (P => R)) = 
begin
[(P => ~~(Q | R));
 [~((P => Q) | (P => R));
  [P;
   ~~(Q | R);
   [Q | R;
    [Q;
     [P;
      Q];
     P => Q;
     ((P => Q) | (P => R));
     F];
    [R;
     [P;
      R];
     P => R;
     ((P => Q) | (P => R));
     F];
    F];
   ~(Q | R);
   F;
   Q];
  (P => Q);
  ((P => Q) | (P => R));
  F];
 ~~((P => Q) | (P => R))];
(P => ~~(Q | R)) => ~~((P => Q) | (P => R));
end;

% 6) ((P => Q) | (Q => P))* 
%  = ~~ ((P* => Q*) | (Q* => P*))
%  = ~~ ((~~P => ~~Q) | (~~Q => ~~P))

proof onewayortheother :  ~~ ((P => Q) | (Q => P)) =
begin
[~ ((P => Q) | (Q => P));
 [P;
  [Q;
   P];
  Q => P;
  ((P => Q) | (Q => P));
  F;
  Q];
 (P => Q);
 ((P => Q) | (Q => P));
 F];
~~ ((P => Q) | (Q => P));
end;