Easier Guide to Formal Logic: Old Symbols

If you're lazy to study formal logic, zero-order logic, first-order logic, second, third, higher, and just want to skip to the goods, then I'll be damned. But it's possible. Today you will learn modal logic, intuitionistic modal logic, deontic and epistemic modal logic, classical rules modal logic, and non-classical modal logic—everything about modal logic even beyond the standard 5 axioms and to the Gödelian-Łukasiewicz, Brouwerian, and Lewisian axioms.

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1. Notation Reference Key

Before building the axiomatic architecture, we establish our structural primitives. Every logical statement is interpreted as an algebraic mapping over a Kripke frame $\mathcal{F} = \langle W, R \rangle$.

Symbol	Structural Definition & Action
$W$	The universal set of all worlds ($w, v, u \in W$).
$A, B \subseteq W$	Propositions, explicitly defined as subsets of $W$.
$\in$	Set membership; denotes a world existing within a proposition's set ($w \in A$).
$\cap, \cup, -$	The classical Boolean operations of intersection, union, and absolute complement (where $-A = W \setminus A$).
$\subseteq$	Global structural validity or containment. The statement $A \subseteq B$ indicates that every world satisfying $A$ unconditionally satisfies $B$.
$-A \cup B$	Local material implication. This represents the set of worlds where either $A$ is false or $B$ is true.
$L(A)$	The structural necessity operator: $L(A) = \{w \in W \mid R(w) \subseteq A\}$.
$M(A)$	The structural possibility operator: $M(A) = \{w \in W \mid R(w) \cap A \neq \emptyset\}$.

2. The Geometric Foundation: Kripke Frames

Instead of treating modal logic as an ungrounded game of shifting syntax, we anchor every statement in a concrete geometric space. We bypass classical truth values ($1$ and $0$) assigned to isolated sentences. Instead, truth is spatial, localized to specific points within a structured universe.

Definition 2.1 (The Kripke Frame). A Kripke Frame is a tuple $\mathcal{F} = \langle W, R \rangle$, where:

• $W$ is a non-empty set whose elements are called worlds (or points, nodes, states).
• $R \subseteq W \times W$ is a binary relation on $W$, known as the accessibility relation. If $(w, v) \in R$, we write $wRv$, meaning "world $v$ is possible from the standpoint of world $w$."

For any given world $w \in W$, we define its successor set $R(w)$ as the collection of all worlds reachable from it:

$$R(w) = \{v \in W \mid wRv\}$$

Definition 2.2 (The Modal Model). A Kripke Model is a tuple $\mathcal{M} = \langle \mathcal{F}, V \rangle = \langle W, R, V \rangle$, where $\mathcal{F}$ is a Kripke frame and $V$ is a valuation function:

$$V: \text{Atomic Propositions} \longrightarrow \mathcal{P}(W)$$

Unlike classical logic, where a valuation assigns an atom to either true or false, here $V$ maps an atomic proposition directly to the set of worlds where it is true.

3. Syntax-Semantics Fusion (The Set-Theoretic Engine)

To eliminate the layers of boilerplate text typical of standard logic presentations, we map both classical and modal operators directly to set operations over the power set of worlds, $\mathcal{P}(W)$.

Let $A, B \subseteq W$ be propositions (represented explicitly as the sets of worlds in which they hold true). The standard classical connectives are defined as follows:

• Conjunction ($\wedge$): $A \cap B = \{w \in W \mid w \in A \text{ and } w \in B\}$
• Disjunction ($\vee$): $A \cup B = \{w \in W \mid w \in A \text{ or } w \in B\}$
• Negation ($\neg$): $-A = W \setminus A = \{w \in W \mid w \notin A\}$
• Strict Implication ($\rightarrow$): $A \subseteq B$

A statement $A \rightarrow B$ is globally valid within a frame if and only if its set containment holds across the entire universe: $A \subseteq B$.

3.1. The Modal Mappings ($L$ and $M$)

We completely discard the traditional square ($\Box$) and diamond ($\diamond$) notation. We use the 1990s structural notation shift, defining the operators as unary set-functions $L$ (Necessity) and $M$ (Possibility):

• Necessity ($L$): The set of all worlds whose successor sets are entirely contained within the target proposition. $$L(A) = \{w \in W \mid R(w) \subseteq A\}$$
• Possibility ($M$): The set of all worlds whose successor sets share at least one element with the target proposition. $$M(A) = \{w \in W \mid R(w) \cap A \neq \emptyset\}$$

3.2. Visualizing the Frame Space

Below is a structural representation of a Kripke frame with three worlds, $W = \{w_1, w_2, w_3\}$, where a proposition $A$ is true only at $w_2$ and $w_3$ ($A = \{w_2, w_3\}$).

4. Algebraic Properties & Duality Proofs

Because $L$ and $M$ are defined strictly as operators on sets, their fundamental properties can be verified directly through set theory without relying on truth tables or lengthy deductive proofs.

Theorem 4.1 (Modal Duality).

For any proposition $A \subseteq W$, the operators $L$ and $M$ are duals:

$$M(A) = -L(-A)$$

Proof: By definition, the negation of a proposition is its complement relative to $W$. Let us expand the right side of the equation step-by-step:

1. Find $-A$: $$-A = \{v \in W \mid v \notin A\}$$
2. Apply the necessity operator $L$ to $-A$: $$L(-A) = \{w \in W \mid R(w) \subseteq -A\}$$ This states that every world accessible from $w$ is not in $A$.
3. Take the complement of the entire set $-L(-A)$: $$-L(-A) = W \setminus \{w \in W \mid R(w) \subseteq -A\}$$
4. The condition $\neg(R(w) \subseteq -A)$ logically means that it is false that all elements of $R(w)$ are outside of $A$. Therefore, there must exist at least one element in $R(w)$ that belongs to $A$: $$\neg(R(w) \subseteq -A) \iff R(w) \cap A \neq \emptyset$$
5. This matches the definition of the possibility operator exactly: $$-L(-A) = \{w \in W \mid R(w) \cap A \neq \emptyset\} = M(A) \quad \blacksquare$$

Theorem 4.2 (Monotonicity of $L$).

If $A \subseteq B$, then $L(A) \subseteq L(B)$.

Proof: Assume $A \subseteq B$ and select an arbitrary world $w \in L(A)$.

1. By definition of $L$, $w \in L(A) \implies R(w) \subseteq A$.
2. By the transitivity of set inclusion, if $R(w) \subseteq A$ and $A \subseteq B$, it follows immediately that $R(w) \subseteq B$.
3. Since $R(w) \subseteq B$, by definition of the necessity operator, $w \in L(B)$.
4. Because any arbitrary $w$ in $L(A)$ is also found in $L(B)$, we have established that $L(A) \subseteq L(B)$. $\quad \blacksquare$

Problem Set 1

Execute these exercises using clean algebraic proofs. Do not use words where set notation can express the same concept.

Problem 1.1: Prove that $L(A \cap B) = L(A) \cap L(B)$ using the definition $L(X) = \{w \in W \mid R(w) \subseteq X\}$.

Problem 1.2: Construct an explicit counter-model (define $W$, $R$, and the subsets $A, B$) to demonstrate that the distribution of possibility over conjunction fails. That is, show: $M(A \cap B) \neq M(A) \cap M(B)$.

Problem 1.3: Prove that $L(A) \cup L(B) \subseteq L(A \cup B)$. Provide a minimal finite frame counter-model proving that the reverse containment $L(A \cup B) \subseteq L(A) \cup L(B)$ is not universally valid.

5. The Distribution Engine: Axiom K

Every normal system of modal logic requires the validity of the distribution principle. Rather than treating this as an arbitrary syntactic rule, we observe that it is an inescapable algebraic trait of the necessity operator $L$ over any arbitrary Kripke frame.

Theorem 5.1 (Axiom K: Distribution).

$$L(-A \cup B) \subseteq (-L(A) \cup L(B))$$

Proof: To prove this without classical truth tables, we translate the set containment into standard element-wise conditions:

1. Let $w \in W$ be an arbitrary world such that $w \in L(-A \cup B)$. By definition of the $L$ operator, this implies: $$R(w) \subseteq (-A \cup B)$$
2. We must show that $w \in (-L(A) \cup L(B))$, which is structurally equivalent to proving that if $w \in L(A)$, then $w \in L(B)$.
3. Assume $w \in L(A)$. By definition, this yields: $$R(w) \subseteq A$$
4. We combine our two assumptions. Every successor $v \in R(w)$ must satisfy two conditions simultaneously: $$v \in A \quad \text{and} \quad v \in (-A \cup B)$$
5. Because $v \in -A \cup B$, $v$ must be an element of either $-A$ or $B$. However, step 4 establishes that $v \in A$ (meaning $v \notin -A$). Therefore, $v$ must belong to $B$.
6. Since every arbitrary successor $v \in R(w)$ is contained within $B$, we have: $$R(w) \subseteq B \implies w \in L(B)$$
7. The conditional holds universally across all worlds, validating the inclusion: $L(-A \cup B) \cap L(A) \subseteq L(B)$, which transposes identically to Axiom K. $\quad \blacksquare$

6. Direct Axiomatization & Frame Correspondence

The power of modal logic emerges when we restrict the geometry of the accessibility relation $R$. Imposing a geometric constraint on $R$ forces a corresponding algebraic inclusion to hold across the power set $\mathcal{P}(W)$.

If a modal axiom is valid on a frame if and only if the frame satisfies a geometric property, we state that the axiom corresponds to that property.

6.1. The Complete Inventory of Core Modal Axioms

Here is the structural mapping of the twelve foundational modal axioms. Every system is constructed by taking the baseline engine $\mathbf{K}$ and adding subsets of these algebraic inclusions.

Axiom	Set Inclusion	Geometric Frame Condition	Formal First-Order Property
K	$L(-A \cup B) \subseteq (-L(A) \cup L(B))$	None (Universal)	Always valid on all frames
D	$L(A) \subseteq M(A)$	Serial	$\forall w \exists v (wRv)$
T	$L(A) \subseteq A$	Reflexive	$\forall w (wRw)$
B	$A \subseteq LM(A)$	Symmetric	$\forall w \forall v (wRv \implies vRw)$
4	$L(A) \subseteq LL(A)$	Transitive	$\forall w \forall v \forall u (wRv \wedge vRu \implies wRu)$
5	$M(A) \subseteq LM(A)$	Euclidean	$\forall w \forall v \forall u (wRv \wedge wRu \implies vRu)$
CD	$M(A) \subseteq L(A)$	Partial Function	$\forall w \forall v \forall u (wRv \wedge wRu \implies v = u)$
G1	$ML(A) \subseteq LM(A)$	Convergent	$\forall w, v, u (wRv \wedge wRu \implies \exists x (vRx \wedge uRx))$
X	$LL(A) \subseteq L(A)$	Dense	$\forall w \forall v (wRv \implies \exists u (wRu \wedge uRv))$
GL	$L(-L(A) \cup A) \subseteq L(A)$	Converse Well-founded	No infinite ascending chains
Grz	$L(-L(-A \cup L(A)) \cup A) \subseteq A$	Modernly Isolated	Partial order with finite paths
McKinsey	$LM(A) \subseteq ML(A)$	Atomicity	$\forall w \exists v (wRv \wedge \forall u (vRu \implies \exists x (uRx \wedge vRx)))$

6.2. Visualizing Geometric Transformations

When a frame satisfies the Confluence (G1) property, any world $w$ that branches out to two distinct paths ($v$ and $u$) must eventually converge at a shared terminal world $x$. This geometry guarantees that if a proposition $A$ is necessary from the standpoint of $v$, its possibility can be seen from $u$.

7. Deep-Dive Mathematical Proofs of Correspondence

To demonstrate how these structural restrictions dictate the containment equations, we explicitly work out the bidirectional proofs for two structural milestones: Axiom 4 and Axiom B.

Theorem 7.1 (Structural Correspondence of Axiom 4).

The algebraic containment $L(A) \subseteq LL(A)$ holds true on a Kripke frame $\mathcal{F}$ if and only if $R$ is transitive.

Proof: Part 1 ($\implies$): We assume the algebraic property $L(A) \subseteq LL(A)$ is universally valid for any valuation on our frame, and we must prove that $R$ is geometrically transitive.

1. Suppose $R$ is not transitive. This means there exist three distinct worlds $w, v, u \in W$ such that $wRv$, $vRu$, but $\neg(wRu)$.
2. We define a targeted assignment for our proposition $A$: let $A = W \setminus \{u\}$.
3. Let us evaluate $w$ relative to $L(A)$. The successor set of $w$ is $R(w)$. Because $\neg(wRu)$, the world $u$ is excluded from $R(w)$. Since $u$ is the only world missing from $A$, it follows that every world in $R(w)$ belongs to $A$. Thus: $$R(w) \subseteq A \implies w \in L(A)$$
4. Next, let us evaluate $w$ relative to $LL(A)$. For $w$ to be an element of $LL(A)$, its successor set $R(w)$ must be entirely contained within $L(A)$. Let's check the successor world $v \in R(w)$:
   • For $v$ to be in $L(A)$, we must have $R(v) \subseteq A$.
   • However, we know that $vRu$, meaning $u \in R(v)$.
   • By our definition in step 2, $u \notin A$. Therefore, $R(v) \nsubseteq A$, which means $v \notin L(A)$.
5. Because $v \in R(w)$ but $v \notin L(A)$, the set inclusion fails: $R(w) \nsubseteq L(A)$. Therefore, $w \notin LL(A)$.
6. We have found a world where $w \in L(A)$ but $w \notin LL(A)$, directly breaking the inclusion rule $L(A) \subseteq LL(A)$. By contradiction, $R$ must be transitive.

Proof: Part 2 ($\impliedby$): We assume $R$ is geometrically transitive, and we must prove the set containment $L(A) \subseteq LL(A)$.

1. Select an arbitrary world $w \in L(A)$. By definition, $R(w) \subseteq A$.
2. To show $w \in LL(A)$, we must show that for any $v \in R(w)$, it follows that $v \in L(A)$ (which means $R(v) \subseteq A$).
3. Select an arbitrary world $v \in R(w)$ and an arbitrary world $u \in R(v)$.
4. Because $wRv$ and $vRu$, the transitivity of $R$ requires that $wRu$. Therefore, $u \in R(w)$.
5. From step 1, we know $R(w) \subseteq A$. Since $u \in R(w)$, it must be that $u \in A$.
6. Because every arbitrary $u \in R(v)$ is an element of $A$, we have $R(v) \subseteq A$, proving $v \in L(A)$.
7. Since every arbitrary $v \in R(w)$ is an element of $L(A)$, we have $R(w) \subseteq L(A)$, proving $w \in LL(A)$. $\quad \blacksquare$

Theorem 7.2 (Structural Correspondence of Axiom B).

The algebraic containment $A \subseteq LM(A)$ holds true on a Kripke frame $\mathcal{F}$ if and only if $R$ is symmetric.

Proof: Part 1 ($\implies$): We assume $A \subseteq LM(A)$ holds universally, and we must show $\forall w, v (wRv \implies vRw)$.

1. Suppose $R$ is not symmetric. There exist worlds $w, v \in W$ such that $wRv$ but $\neg(vRw)$.
2. Define the proposition $A$ specifically as the singleton set containing only $w$: $A = \{w\}$.
3. By definition, since $w \in A$, our world immediately satisfies the left-hand side of our axiom.
4. Now examine the right-hand side, $LM(A)$. For $w$ to be an element of $LM(A)$, its successor set must be contained within $M(A)$, meaning $R(w) \subseteq M(A)$. We check the accessible world $v \in R(w)$:
   • For $v$ to be in $M(A)$, its successor set must intersect with $A$: $R(v) \cap A \neq \emptyset$.
   • Since $A = \{w\}$, this requires $w \in R(v)$, which is equivalent to saying $vRw$.
   • However, step 1 specifies that $\neg(vRw)$. Therefore, $R(v) \cap A = \emptyset$, meaning $v \notin M(A)$.
5. Because $v \in R(w)$ but $v \notin M(A)$, it is false that $R(w) \subseteq M(A)$. Thus, $w \notin LM(A)$.
6. We have $w \in A$ alongside $w \notin LM(A)$, violating the structural containment. By contradiction, $R$ must be symmetric.

Proof: Part 2 ($\impliedby$): We assume $R$ is symmetric, and we must prove $A \subseteq LM(A)$.

1. Let $w$ be an arbitrary world in $A$.
2. To prove $w \in LM(A)$, we must confirm that every successor $v \in R(w)$ belongs to $M(A)$.
3. Select an arbitrary successor $v \in R(w)$.
4. Since $R$ is symmetric, $wRv$ implies $vRw$. Therefore, $w \in R(v)$.
5. Since we established in step 1 that $w \in A$, it follows that $w \in R(v) \cap A$. This means the intersection is non-empty: $R(v) \cap A \neq \emptyset$.
6. By definition of possibility, this non-empty intersection means $v \in M(A)$.
7. Because every arbitrary successor $v \in R(w)$ is found within $M(A)$, we have $R(w) \subseteq M(A)$, which proves $w \in LM(A)$. $\quad \blacksquare$

Problem Set 2

Problem 2.1: Prove the structural correspondence for Axiom D ($L(A) \subseteq M(A)$). Show explicitly how a violation of frame seriality ($\exists w \text{ s.t. } R(w) = \emptyset$) invalidates this containment by turning the necessity operator into a vacuous universal quantifier.

Problem 2.2: Using step-by-step set algebraic manipulation, prove that the combination of Axiom T ($L(A) \subseteq A$) and Axiom 5 ($M(A) \subseteq LM(A)$) algebraically forces the validation of Axiom B ($A \subseteq LM(A)$). Do not invoke frame properties; complete this proof entirely through syntactic-algebraic manipulation of the set operations.

Problem 2.3: The Density Axiom X states that $LL(A) \subseteq L(A)$. Construct an explicit three-world counter-model frame ($W = \{w_1, w_2, w_3\}$) that breaks density, and show the exact subset assignment for $A$ that refutes the inclusion.

8. Applied Spaces: Deontic, Epistemic, and Intuitionistic Mappings

To adapt our set-theoretic modal engine to specific philosophical and mathematical domains, we re-map the semantic interpretations of $L$ and $M$ while retaining the underlying frame architectures.

8.1. Deontic ($O / P$) and Epistemic ($K / P$) Systems

• Deontic Logic (Norms and Actions): We map $L \to O$ (Obligation) and $M \to P$ (Permission).
• The Baseline Constraint: Deontic frameworks require System D ($O(A) \subseteq P(A)$). A frame where $R(w) = \emptyset$ (a dead-end world) would validate $O(A)$ vacuously while invalidating $P(A)$, signaling a broken norm system.
• Epistemic Logic (Knowledge and Belief): We map $L \to K$ (Knowledge: "It is known that...") or $L \to B$ (Belief).
• Positive Introspection (Axiom 4): $K(A) \subseteq KK(A)$ (If you know $A$, you know that you know $A$).
• Negative Introspection (Axiom 5): $-K(A) \subseteq K(-K(A))$ (If you do not know $A$, you know that you do not know it).

8.2. Intuitionistic Modal Logic (Constructive Topology)

To construct a non-classical modal space, we drop the classical absolute complement law, meaning the Law of Excluded Middle fails globally:

$$-A \cup A \neq W$$

An Intuitionistic Kripke frame forces a dual-relation structure $\langle W, \le, R \rangle$, where $\le$ is a partial order representing the growth of information state, and $R$ is the standard modal accessibility.

• Monotonicity Constraint: If $w \in A$ and $w \le v$, then $v \in A$ (Truth is persistent; once verified, it cannot be lost).
• Intuitionistic Necessity: $L(A) = \{w \in W \mid \forall v \ge w : R(v) \subseteq A\}$.

9. Heavy Metal Axiomatics & Syntactic Reduction

When we push modal logic to its structural limits, the interactions between deeply nested operators collapse into clean, predictable algebraic patterns.

9.1. Lewisian Systems (S1–S3)

Before modern relational semantics, Clarence Irving Lewis defined modal spaces using Strict Implication ($\prec$) to eliminate material implication paradoxes. We define strict implication set-theoretically as:

$$A \prec B \equiv L(-A \cup B)$$

Systems S1, S2, and S3 are built on weak, non-normal frame foundations where the distribution rule (Axiom K) does not hold universally. They rely instead on strict matrices that govern strings of nested implications.

9.2. The Gödel-Löb Provability Engine ($\mathbf{GL}$)

In mathematical logic, we map $L(A)$ to mean "A is provable in Peano Arithmetic (PA)". This yields System GL, governed by the Löb Axiom:

$$\mathbf{GL: } \quad L(-L(A) \cup A) \subseteq L(A)$$

• Geometric Boundary: A Kripke frame satisfies the $\mathbf{GL}$ inclusion if and only if $R$ is transitive and converse well-founded (it contains no infinite ascending chains of accessibility: $w_1 R w_2 R w_3 R \dots$). This geometry mirrors Gödel's Second Incompleteness Theorem within spatial structures.

9.3. The S5 Syntactic Collapse

In System S5 (where $R$ is reflexive, symmetric, and transitive, forming an equivalence relation), the distinction between local accessibility and global necessity disappears. This causes every complex string of nested modal operators to collapse into its final operator.

$$\begin{align*} LL(A) &= L(A) \\ MM(A) &= M(A) \\ LM(A) &= M(A) \\ ML(A) &= L(A) \end{align*}$$

Theorem 9.1 (The S5 Reduction Theorem).

Any arbitrary sequence of modal operators $\mathcal{O}_1 \mathcal{O}_2 \dots \mathcal{O}_n$ (where $\mathcal{O}_i \in \{L, M\}$) reduces identically to $\mathcal{O}_n$.

$$M L M L M L L M(A) \equiv M(A)$$

Final Review Problems

Problem 3.1: Prove that in an S5 model, if $w \in M(L(A))$, then $w \in L(A)$. Use the geometric properties of an equivalence relation ($R$ is reflexive, symmetric, and transitive) to show this set containment.

Problem 3.2: In epistemic logic, the puzzle of the "Fitch Paradox" shows that if all truths are knowable ($A \subseteq M(K(A))$), then all truths are actually known ($A \subseteq K(A)$).

Using the explicit substitution $A \cap -K(A)$ into the knowability thesis, prove algebraically that this blindspot proposition collapses into the empty set ($\emptyset$), thereby demonstrating that unknown truths cannot be known.