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7.5  Predicate tautologies


Definition 7.5.1

A predicate calculus formula α is called a tautology (or the law of the functional calculus) if its value is true regardless of the valuation of the variables and the interpretation of functional and relational symbols which occur in α, i.e. for every structure STR and for every valuation v of variables in the domaine of  STR we have    STR,v|= α.

Instead of  " α is a tautology of predicate calculus" we write shortly |= α. The set of all predicate tautologies can be characterised as the intersection of the sets of formulas valid in an arbitrary structure in which the language of predicate calculus is interpreted. 

The tautologies of the propositional calculus form the set of "patterns" for tautologies of predicate calculus. The following lemma makes the above statement precise.

Lemma 7.5.1

If α is a propositional calculus tautology, then the result of substituting predicate formulas for all propositional variables in α is a predicate calculus tautology.

We do not derive detailed proof of the above-mentioned lemma. We only remind the Reader that if α is a propositional calculus tautology, then, regardless of propositional variables' values, the value of the formula is true.

Example 7.5.1

Let p and q be the propositional variables and α(x,y) a formula of predicate calculus with two free variables x and y.
(a) The formula (p ∨ ¬p) is a tautology of propositional calculus, thus (according to lemma 7.5.1) the formulas

(x≤y ∨ ¬ x≤y),   (( ∃y)(x + y >z) ∨ ¬ (∃y)(x + y >z)),   ((∀x) (∃ y) α(x,y) ∨  ¬ (∀ x)(∃ y) α(x,y))

are tautologies of predicate calculus.
However, the formula  (x≤y ∨ x>y) is not tautology because the interpretation of the symbol  ≤ does not have to be the completion of the  interpretation of  >. The formula ¬ (x ≤y) over the set of real numbers is equivalent to the formula (x > y), but with a different interpretation of symbols ≤ and > in a different structure it may not be true. In consequence, there are structures in which the formula (x≤y ∨ x>y)  is false.
(b) The formula (¬p →  (p → q)) is a propositional tautology, hence the formula

(¬ 2|x →  (2|x → 3|x))

is the predicate calculus tautology, and, therefore, it is valid, in particular, in the struture of natural numbers.


The predicate calculus tautologies are schemas of propositional functions which are valid in any structure, i.e true for all valuations of variables in the domain of the structure. The validity of these formulas is not due to the assumed meaning of non-logical symbols or due to the accepted variable's valuation, but it is a consequence of the structure of the formula itself. The following lemma presents a few examples of the predicate calculus tautologies:

Lemma 7.5.2

For any formulas α(x) and β(x) the following formulas are tautologies of predicate calculus:

  1. ( ∀x) α(x) → ( ∃x)α(x)
  2. (¬ ( ∀x) α(x)) → (( ∃x) ¬ α(x))
  3. ( ∀x) (( α(x) ∧ β(x)) ↔ (( ∀x) α(x) ∧ ( ∀x) β(x))
  4. ( ∃x)( α(x) ∨ β(x)) ↔ (( ∃x)α(x) ∨ ( ∃x) β(x))

Proof.

Let STR be an arbitrary structure with a non-empty domain X and let v be a valuation in STR.
Ad (1)  If the implication  ( ∀x)α(x) → ( ∃x)α(x)  is false for some interpretation of the formula α(x) over the structure STR, then {x ∈ X: α(x)} = X as well as {x ∈ X: α(x)} = ∅. That is X = ∅, a contradiction.

Ad (2) If STR,v|= ¬(∀x) α(x), then the valuation v does not satisfy the formula ( ∀x) α(x). It means that the formula α(x) is false for some value of x in STR. Let us denote such value of x by b. Hence,   STR, v(b/x)|= ¬ α(x). Thus, according to the definition of the existential quantifier semantics  we have STR,v |= (( ∃x) ¬ α(x)).

The proof of the remaining two equivalences are left to the Reader.

Remark: The second tautology mentioned in  lemma 7.5.2  is called  de Morgan's law for quantifiers. There is dual de Morgan's law which is read as (¬ (∃x) α(x)) → ((∀x) ¬ α(x)).

Example 7.5.2

  1. The formula (¬(∀x) (x2 > x+1) → ( ∃x)¬ (x2 > x+1)) is a particular case of de Morgan's law, thus it is a true proposition over the structure of real numbers R.
  2. In the structure of natural numbers the following proposition is valid ( ∀x)(0≤x ∧ ( ∃y)(x≤y)). Therefore, in accordance with the tautology (3) from lemma 7.5.2, the formula (( ∀x)(0≤x) ∧ (∀x)(∃y) (x≤y)) is  valid in N.

Question 7.5.1 Prove that for any formula α(x) and any formula β in which x is not a free variable the formula ( ∃x)( α(x) ∧ β) ↔ (( ∃x) α(x) ∧ β) is a tautology of predicate calculus (law of inclusion and exclusion of existential quantifier).


----- Check the answer -----

Question 7.5.2 Which of the following formulas are predicate calculus tautologies?

  1. ¬ ( ∃x) α(x) → ( ∀x) ¬ α(x)
  2. ( ∀x) ( α(x) ∨ β(x)) → (( ∀x) α(x) ∨ ( ∀x) β(x))
  3. ( ∃x) ( α(x) ∧ β(x)) → (( ∃x) α(x) ∧ ( ∃x) β(x))
  4. (( ∀x) α(x) ∨ ( ∀x) β(x)) → ( ∀x) ( α(x) ∨ β(x))
  5. (( ∃x) α(x) ∧ ( ∃x) β(x)) → ( ∃x) ( α(x) ∧ β(x))


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