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6. First-Order Logic

We extend our propositions and our linear proof format to include universal and existential quantification. Here is an example demonstrating All-introduction and -elimination and Exists-introduction:
 
proof AllEx : !y:t. (!x:t. A(x)) => ?x:t. A(x) =
begin
    [ c : t;
      [ !x:t. A(x);
        A(c);
        ?x:t. A(x) ];
      (!x:t. A(x)) => ?x:t. A(x)];
    !y:t. (!x:t. A(x)) => ?x:t. A(x)
end;
The scope of a quantification starts at the `.' and extends as far to the right as syntactically possible. Thus the All-quantified variable `y' is bound in the whole proposition, whereas the scope of `!x:t.' is limited by parentheses to preserve the intended meaning of the proposition. Invoking the proof checker Tutch, it gives the following justifications
 
Proving AllEx: !y:t. (!x:t. A x) => ?x:t. A x ...
  1  [ c: t;
  2    [ !x:t. A x;
  3      A c;                                          by ForallE 2 1
  4      ?x:t. A x ];                                  by ExistsI 1 3
  5    (!x:t. A x) => ?x:t. A x ];                     by ImpI 4
  6  !y:t. (!x:t. A x) => ?x:t. A x                    by ForallI 5
QED   
Universal quantification !y:t. B(y) can be introduced by a frame [c:t; ... B(c)] as we see in the last line: Here B(y) is
 
(!x:t. A x) => ?x:t. A x
which does not contain any occurrence of y. Universal quantification can be eliminated if we have a term of the type over which the quantification is ranging. An example can be seen in line 3: We have an All-quantified assertion !x:t. A x in line 2 and a term c : t, the parameter introduced in line 1. Thus we can deduce A x where all occurrences of x are replaced by c, which is A c.

To introduce an existential quantification ?x:t. A x (line 4) we need a witness c : t (line 1) and a proof for the special instance of the proposition A c (line 3). For existential elimination we consider the following example:
 
proof ExNotImpNotAll : (?x:t. ~A(x)) => ~!x:t. A(x) =
begin
[ ?x:t. ~A(x);
  [ !x:t. A(x);
    [ c: t, ~A(c);
      A(c);
      F ];
    F ];
  ~!x:t. A(x) ];
(?x:t. ~A(x)) => ~!x:t. A(x); 
end;
To prove falsehood F in proof line 6, we eliminate the existential quantification in the first line. This gives us two new hypotheses to show our goal: A witness c : t and a proof of ~A(c). In principle Exists-elimination is used in the same way as disjunction elimination. Note that we extended our frame syntax to include the introduction of several hypotheses, separated by commas.

For further information consult the reference.


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