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 March 20th, 2014, 08:05 AM #1 Senior Member   Joined: Apr 2010 Posts: 451 Thanks: 1 wf formulas Which of the following are well formed formulas: $\forall x[(\forall y(xy=y))\Longrightarrow x=1]$ $\exists 1[ a.1=1.a=a]$ $\exists 1\forall a[1.a=a.1=a]$
 March 20th, 2014, 12:19 PM #2 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 Re: wf formulas For one, a quantifier must be followed by a variable. Do you have the definition of a wff? Have you read any examples of applying this definition?
March 20th, 2014, 04:24 PM   #3
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Re: wf formulas

Quote:
 Originally Posted by Evgeny.Makarov For one, a quantifier must be followed by a variable. Do you have the definition of a wff? Have you read any examples of applying this definition?
can't you phrase the thing with the symbol 1 as a variable, stating $exists 1\forall a\st 1\cdot a=a\cdot 1=a$, there is no conflict. Then when you show that 1 is unique (this requires proof) it can be treated as a constant

March 24th, 2014, 01:18 PM   #4
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Re: wf formulas

Quote:
 Originally Posted by outsos can't you phrase the thing with the symbol 1 as a variable
Well, is the symbol 1 a variable or a constant? The definition of wffs starts with fixing a vocabulary, which consists of several types of symbols: objects variables, propositional symbols, constants, auxiliary symbols such as parentheses and commas and so on. Once this is fixed, you can't treat a constant as a variable, at least in the standard definition of wffs.

Quote:
 Originally Posted by outsos Then when you show that 1 is unique (this requires proof) it can be treated as a constant
There is indeed a way to introduce new constants and functional symbols. For a constant, you need to prove that, for some property P(x), there exists an object satisfying P(x) and it is unique. Then you can give a name to this object. After that you have a procedure that translates any formula with this name into an equivalent formula that does not use the name and only uses P(x). This procedure works on the metalevel, i.e., the procedure itself and its properties are described in English rather than by formulas of first-order logic. Also, the definition of formulas does not change, but one gets a new vocabulary with an additional constant symbol.

To sum up: the definition of a well-formed formula depends on the vocabulary. If in the given vocabulary 1 is a constant, then it cannot immediately follow a quantifier.

April 24th, 2014, 04:23 PM   #5
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Quote:
 Originally Posted by Evgeny.Makarov Well, is the symbol 1 a variable or a constant? The definition of wffs starts with fixing a vocabulary, which consists of several types of symbols: objects variables, propositional symbols, constants, auxiliary symbols such as parentheses and commas and so on. Once this is fixed, you can't treat a constant as a variable, at least in the standard definition of wffs. There is indeed a way to introduce new constants and functional symbols. For a constant, you need to prove that, for some property P(x), there exists an object satisfying P(x) and it is unique. Then you can give a name to this object. After that you have a procedure that translates any formula with this name into an equivalent formula that does not use the name and only uses P(x). This procedure works on the metalevel, i.e., the procedure itself and its properties are described in English rather than by formulas of first-order logic. Also, the definition of formulas does not change, but one gets a new vocabulary with an additional constant symbol. To sum up: the definition of a well-formed formula depends on the vocabulary. If in the given vocabulary 1 is a constant, then it cannot immediately follow a quantifier.
Why the rules of wffs are set in a such a way that we cannot quantify a constant.

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