Questions: Interpretation, Truth, and Satisfaction of Formulas

Question 1

Consider the formula 'x > 0' evaluated in the structure M = (ℝ, <). What is the truth value of this formula in M?

• A. True, because most real numbers are positive
• B. False, because x could be assigned a negative value
• C. It has no truth value in M without specifying a variable assignment for x
• D. True, because ℝ is an ordered field and positivity is well-defined

Answer: (C)

The formula 'x > 0' contains a free variable x. Free variables do not have truth values in isolation — truth requires either a quantifier binding x or a variable assignment s specifying which element of ℝ the variable x refers to. Under assignment s with s(x) = 3, the formula is true. Under s with s(x) = −5, it is false. The formula is neither 'true in M' nor 'false in M' because it depends on what x refers to. Option A misapplies a probabilistic intuition. Option B correctly identifies that x could be negative but draws the wrong conclusion — the formula isn't false, it's indeterminate without an assignment.

Question 2

The formula ∀x P(x) is evaluated in a structure M with domain D = {1, 2, 3} and P^M = {1, 2}. What is the truth value?

• A. True, because most elements of D satisfy P
• B. False, because element 3 is not in P^M, so P(3) fails
• C. True, because ∀x is a universal quantifier and D is finite
• D. Undefined, because P is not total over D

Answer: (B)

∀x P(x) is satisfied in M if and only if P(d) is true for every element d in the domain D. Since D = {1, 2, 3} and P^M = {1, 2}, we must check P(1), P(2), and P(3). P(1) and P(2) are true (1 and 2 are in P^M), but P(3) is false (3 is not in P^M). Because ∀x demands the formula hold for all elements and it fails at 3, the universal statement is false. Option A applies a majority rule that is not the definition of universal quantification. Option C introduces an irrelevant special case about finite domains. Option D is wrong: predicates are total by definition — P^M is a subset of D, so every element either is or is not in P^M.

Question 3

A closed formula (one with no free variables) has a definite truth value — either true or false — in any given structure, without specifying a variable assignment.

• T. True
• F. False

Answer: True

A closed formula has no free variables; all variables are bound by quantifiers. Since there is nothing left unspecified, the satisfaction relation M ⊨ φ (without mentioning any assignment s) is well-defined. You can evaluate it by tracing through the recursive satisfaction clauses: for each quantifier, check all elements of the domain; for atomic formulas, check whether the tuples are in the predicate extension. The result is a definite true or false. This is why sentences (closed formulas) in first-order logic are the objects that can be true or false 'in a model,' while open formulas with free variables are only satisfied or not satisfied relative to variable assignments.

Question 4

The truth value of a formula in first-order logic is determined by its syntactic structure alone, without reference to any particular interpretation or variable assignment.

• T. True
• F. False

Answer: False

This confuses syntax with semantics. Syntactic structure determines whether a string is a well-formed formula, but truth values are entirely semantic — they depend on which interpretation is being used. The formula P(a) could be true in one structure (where P^M contains the element assigned to a) and false in another (where it does not). Even a formula like '∀x (x = x)' which is logically valid (true in every interpretation) gets its truth value from the semantic rule that identity is reflexive in every structure — this is a semantic fact, not a syntactic one. In first-order logic, there is no truth without interpretation.

Question 5

Why can't we assign a truth value to a formula with free variables without specifying a variable assignment, and how does this differ from propositional logic?

Model answer: In first-order logic, a formula like 'P(x)' contains the free variable x, which acts as a placeholder that could refer to any element of the domain. Without a variable assignment specifying which element x refers to, the formula has no definite meaning — it might be true of some elements and false of others. A variable assignment s maps each free variable to a specific domain element, completing the interpretation and enabling truth evaluation. In propositional logic, this issue doesn't arise: sentence letters like P and Q are already zero-ary (they take no arguments and refer to no objects), so a truth assignment directly gives them a truth value with no further specification needed. First-order logic's expressive power — the ability to quantify over objects and express properties of arbitrary domain elements — is exactly what creates this dependence on variable assignments for open formulas.

The key contrast is between propositional sentence letters (complete bearers of truth values under a truth assignment) and first-order predicate formulas with free variables (incomplete bearers requiring a variable assignment to specify what the variables denote). The answer should explain what free variables are (placeholders), why they create the dependence (they could refer to any domain element), and why propositional logic avoids this (sentence letters have no arguments).