Question

Show that \(\forall x(\varphi \vee \psi(x)) \rightarrow(\varphi \vee \forall x \psi(x))(x \notin F V(\varphi))\) holds in all Kripke models with constant domain function (i.e. \(\forall k l(D(k)=D(l))\).

Short answer

The implication holds due to the constant domain and free variable conditions.

Step-by-step solution

Understanding the Statement

The statement we need to prove is \(\forall x(\varphi \vee \psi(x)) \rightarrow (\varphi \vee \forall x \psi(x))\), where \( x otin FV(\varphi)\). This means \( x \) does not appear freely in \( \varphi \). Our goal is to prove that if for each \( x \), the disjunction \(\varphi \vee \psi(x)\) holds, then the disjunction \(\varphi \vee \forall x \psi(x)\) holds in a Kripke model.

Setting up the Kripke Model

A Kripke model consists of a set of worlds, a relation between those worlds, and a valuation function that assigns truth values to each atomic proposition at each world. In our scenario, the domain is constant across all worlds, meaning \(D(k)=D(l)\) for any worlds \(k\) and \(l\). This means all individuals exist in every world.

Analyzing the Antecedent

The formula \(\forall x(\varphi \vee \psi(x))\) is true at a world \(w\) if, for every element \(a\) in the domain \(D(w)\), the formula \(\varphi \vee \psi(x/a)\) is satisfied at \(w\). So, at every world \(w\), either \(\varphi\) is true, or \(\psi(x/a)\) is true for each \(a\).

Analyzing the Consequent

The consequent \(\varphi \vee \forall x \psi(x)\) holds at a world \(w\) if either \(\varphi\) is true or \(\forall x \psi(x)\) is true at \(w\). Given the constant domain assumption, \(\forall x \psi(x)\) is true if \(\psi(x/a)\) is true for every element \(a\) in the domain of any world (which is constant).

Proving the Implication

Assume \(\forall x(\varphi \vee \psi(x))\) holds at \(w\). For every \(a\), \(\varphi \vee \psi(x/a)\) holds. If \(\varphi\) is true, \(\varphi \vee \forall x \psi(x)\) holds by the disjunction. Otherwise, if \(\varphi\) is not true, \(\psi(x/a)\) must be true for every \(a\) due to the constant domain, ensuring \(\forall x \psi(x)\) is true. Hence, \(\varphi \vee \forall x \psi(x)\) holds.

Key concepts

Constant Domain

In Kripke models, a constant domain signifies that the same set of individuals is present in every possible world within the model. This greatly simplifies reasoning about such models because the properties of entities don't change as you move from one world to another. For example, let's consider two worlds, \(k\) and \(l\). In a model with a constant domain, \(D(k) = D(l)\), meaning every entity that exists in world \(k\) also exists in world \(l\). This characteristic is particularly important in proofs involving quantifiers, as it allows us to assume that any statement involving individuals is universally applicable across all worlds in the model.
Having a constant domain also means that when assessing whether a statement such as \(\forall x \psi(x)\) is true, we don't need to worry about domain variations. We simply verify that each \(\psi(x/a)\) is satisfied for every individual \(a\) across the entire set of worlds.

Free Variables

Free variables are variables in a logical formula that are not bound by any quantifier within that formula. They stand out because their truth values depend on the specific assignment of values they receive. For instance, in a formula like \(\psi(x)\), \(x\) is a free variable if it's not governed by a quantifier such as \(\forall\) or \(\exists\). Why does this matter?
In the context of the solution, it is given that \(x otin FV(\varphi)\) where \(FV(\varphi)\) represents the set of free variables in \(\varphi\). This means \(x\) doesn't impact \(\varphi\), simplifying our analysis. The absence of \(x\) as a free variable in \(\varphi\) means that \(\varphi\) is evaluated independently of \(x\), and thus, assumptions like \(\forall x (\varphi \vee \psi(x))\) focus mainly on the behavior of \(\psi(x)\) rather than \(\varphi\).

Disjunction

Disjunction is a logical operation often expressed with the symbol \(\vee\), which corresponds to an 'or' in natural language. It is a fundamental concept where a statement \(\varphi \vee \psi\) is true if at least one of the components \(\varphi\) or \(\psi\) is true. This principle underlies many logical arguments and flows naturally when reasoning about such compositions.
In the problem statement, disjunction appears as \(\varphi \vee \psi(x)\). Here, the expression indicates that at any world within the Kripke model, it is sufficient for either \(\varphi\) to be true, \(\psi(x)\) to be true, or both, to satisfy the disjunction. When dealing with implications in logic, understanding how disjunction unfolds is crucial. For instance, in the Kripke model, if \(\forall x(\varphi \vee \psi(x))\) holds, each possible world and each element in that world make the statement true if at least one of the disjuncts is satisfied.
This is key because it influences the movement from \(\forall x (\varphi \vee \psi(x))\) to \(\varphi \vee \forall x \psi(x)\). Thus, mastering disjunction allows clearer predictions of logical outcomes.

Universal Quantification

Universal quantification, symbolized by \(\forall\), asserts that a particular property or statement applies to all elements within a certain domain. In a formula like \(\forall x \psi(x)\), it indicates that \(\psi(x)\) holds true no matter which individual \(x\) is considered.
This concept is a keystone in many logical systems, including Kripke models. It ensures that propositions are uniformly true across various scenarios or "worlds,” a pivotal feature when domains are constant.
In analyzing the antecedent \(\forall x(\varphi \vee \psi(x))\) in the exercise, universal quantification means examining each member of the domain to check whether \(\varphi \vee \psi(x/a)\) is satisfied. The constant domain condition simplifies this verification, as the domain doesn’t fluctuate between worlds. Similarly, the understanding that \(\forall x \psi(x)\) requires \(\psi(x)\) to be true for every member of the domain across all possible universes in the model is simplified by the model's constant domain.
Grasping how universal quantification operates in logic allows for precision in evaluating complex statements, especially those involving nested or composite quantifiers.