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Each Kripke model with bottom node \(k_{0}\) can be turned into a model over a tree as follows: \(K_{t r}\) consists of all finite increasing sequences \(\left\langle k_{0}, k_{1}, \ldots, k_{n}\right\rangle, k_{i}

Short Answer

Expert verified
Sequences in \(K_{tr}\) satisfy \(\varphi\) iff the final node \(k_n\) satisfies \(\varphi\).

Step by step solution

01

Understand the Kripke Model Components

Begin by understanding that a Kripke model consists of a set of nodes/possible worlds and a relation that indicates how these worlds are connected. Here, \(k_0\) is the bottom node, and each \(k_{n}\) can be transitioned to \(k_{n+1}\) if \(k_{n+1}\) is accessible from \(k_{n}\).
02

Define Finite Increasing Sequences

The model over the tree, \(K_{tr}\), contains sequences \(\langle k_0, k_1, \ldots, k_n \rangle\) where each component is reachable, specifically \(k_{i} < k_{i+1}\). These sequences represent paths in the tree model.
03

Understand Evaluation in the Tree Model

The tree model evaluation \(\mathfrak{A}_{tr}(\langle k_0, \ldots, k_n \rangle)\) corresponds to \(\mathfrak{A}(k_n)\) from the original model. This means the assignment of logical values at the node \(k_n\) is used to evaluate formulas at the sequence-pair level in \(K_{tr}\).
04

Relate Forcing in Tree Model to Original Model

We need to show that a sequence \(\langle k_0, \ldots, k_n \rangle\) satisfies a formula \(\varphi\) in \(K_{tr}\) if and only if the node \(k_n\) satisfies \(\varphi\) in the original model, i.e., \(\langle k_0, \ldots, k_n \rangle \models_{tr} \varphi \Leftrightarrow k_n \Vdash \varphi\). This happens because the truth value in the sequence is precisely the truth value at the final node \(k_n\), as assigned by \(\mathfrak{A}_{tr}\).
05

Conclusion

Since each sequence evaluation mirrors the evaluation at the last node \(k_n\), the forcing condition holds. Therefore, forcing in the tree model \(\Vdash_{tr}\) corresponds directly to forcing in the original model at \(k_n\): \(\langle k_0, \ldots, k_n \rangle \models_{tr} \varphi\) is true if and only if \(k_n \Vdash \varphi\) is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite Increasing Sequences
In the realm of Kripke models, finite increasing sequences are vital for understanding how different states or nodes relate to each other over a tree structure. Imagine a sequence as a path of logical steps connecting various points along a tree. Each point or node in this path is denoted as \(k_i\), with the sequence beginning at \(k_0\) and rising through intermediate points all the way to \(k_n\).

A finite increasing sequence,
  • Starts from the bottom node \(k_{0}\),
  • Progresses through nodes such that \(k_i < k_{i+1}\),
  • Continues until reaching the node \(k_n\).
These sequences play a crucial role in transitioning from the Kripke model to a tree model, where each step in the sequence signifies an accessible move from one world or state to the next.
Tree Model
The tree model depiction within Kripke semantics transforms our understanding of logical worlds by representing them as branches of a tree. Each branch, or path, in the tree model is constructed from the finite increasing sequences we discussed earlier. This model allows us to visually and logically explore how various worlds interconnect.

In this model, the set \(K_{tr}\) consists of sequences such as \(\langle k_0, k_1, \ldots, k_n \rangle\). Each sequence demonstrates a path through the tree, where each node \(k_i\) signifies a possible state within this logical framework.

To think of it another way, a tree model can be seen as a map of all possible logical developments, with each path representing a different potential progression through logical states.
Forcing Relation
The concept of a forcing relation is integral in linking the tree model to logical evaluation. The forcing relation, denoted \(\Vdash_{tr}\), helps us determine when a particular sequence of nodes satisfies certain logical formulas. This relation is essential for projecting the properties of the Kripke model onto its tree model counterpart.

The main principle of the forcing relation is as follows:
  • A sequence \(\langle k_0, \ldots, k_n \rangle\) supports a formula \(\varphi\) in \(K_{tr}\) if \(k_n\) satisfies \(\varphi\) in the original Kripke model.
This connection shows that the truth of a formula in the tree model depends on its truth in the original model at the endpoint of the sequence.
Logical Evaluation
Logical evaluation in the context of tree models is an assessment of truth values at various nodes based on their respective assignments. The evaluation function \(\mathfrak{A}_{tr}(\langle k_0, \ldots, k_n \rangle)\) matches \(\mathfrak{A}(k_n)\), meaning the logical truth at the sequence level in \(K_{tr}\) is directly derived from the node \(k_n\) in the Kripke model.

This illustrates an important connection:
  • The value assigned to a sequence in the tree is precisely that of its final node \(k_n\).
  • Logical formulas remain true across both models consistently.
Through logical evaluation, we ensure the correspondence between tree paths and their original nodes, confirming that a formula is true along a path if it holds at its last point.

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Most popular questions from this chapter

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. \(\forall k l(D(k)=D(l))\).

Consider intuitionistic predicate logic without function symbols. Prove the following extension of the existence property: \(\vdash \exists y \varphi\left(x_{1}, \ldots, x_{n}, y\right) \Leftrightarrow\) \(\vdash \varphi\left(x_{1}, \ldots, x_{n}, t\right)\), where \(t\) is a constant or one of the variables \(x_{1}, \ldots, x_{n} .\) (Hint: replace \(x_{1}, \ldots, x_{n}\) by new constants \(a_{1}, \ldots, a_{n}\) ).

Let \(\mathbf{D}=\mathbf{R}[X] / X^{2}\) be the ring of dual numbers. D has a unique maximal ideal, generated by \(X\). Consider a Kripke model with two nodes \(k_{0}, k_{1} ; k_{0}

Consider a language with identity and function symbols, and interpret a \(n\)-ary symbol \(F\) by a function \(F_{k}: D(k)^{n} \rightarrow D(k)\) for each \(k\) in a given Kripke model \(\mathcal{K}\). We require monotonicity: \(k \leq l \Rightarrow F_{k} \subseteq F_{l}\) and preservation of equality, where \(a \sim_{k} b \Leftrightarrow k \Vdash \bar{a}=\bar{b}_{1}: \vec{a} \sim_{k} \vec{b} \Rightarrow F_{k}(\vec{a}) \sim_{k}\) \(F_{k}(\vec{b})\). (i) Show \(K \Vdash \forall \vec{x} \exists ! y(F(\vec{x})=y)\) (ii) Show \(\mathcal{K} \| I_{4}\). (iii) Let \(\mathcal{K} \|-\forall \vec{x} \exists ! y \varphi(\vec{x}, y)\), show that we can define for each \(k\) and \(F_{k}\) satisfying the above requirements such that \(\mathcal{K} \| \forall(\vec{x} \varphi(\vec{x}, F(\vec{x}))\). (iv) Show that one can conservatively add definable Skolem functions. Note that we have shown how to introduce functions in Kripke models, when they are given by "functional" relations. So, strictly speaking, Kripke models with just relations are good enough.

Show that HA \(\vdash \varphi \vee \psi \leftrightarrow \exists x(x=0 \rightarrow \varphi) \wedge(x \neq 0 \rightarrow \psi))\),

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