LOGIC Course


Mathematical logic is useful in many ways. First, it led to globalize theory: in several disciplines linked to mathematics or IT, if you want to prove a hypothesis or develop it, you (have to) use logic.

In our case, it enables us to verify properties or algorithms. By the way, logic is well-known through automated theorem proving, expert systems, or proof assistants. For instance, the four-color theorem (cf graph) which states that only four colors are needed to color the regions of any map so that no two adjacent (nearby) regions have the same color was proved by a computer using logic.


You will start with basic knowledge such as Structural induction (induction), Semantics of logic (Boolean algebra, Truth table, ...), and a simple version of natural deduction. Then, you'll head for the centerpiece: First-order logic.

Structural induction

$\mathcal{B}$ is a set of constant symbols (ensemble de bases)
$\mathcal{K}$ is a set of function symbols (ensemble d'opérations)

So <$\mathcal{B},\mathcal{K}$> is a set created by induction, the smallest set E satisfying the following criteria:

  • $\mathcal{B} \subset E$
  • $f$ an $n_{f}$-ary function symbol and $x_{1}, \cdots, x_{n_{f}} \in E$, then $f(x_{1}, \cdots, x_{n_{f}}) \in E$ too

Proof by induction: Let's consider a hypothesis $P$.
To demonstrate that all the elements of <$\mathcal{B},\mathcal{K}$> satisfy $P$, you have just to verify that:

  • $\forall x \in \mathcal{B}$, $P(x)$ is satisfied
  • $\forall f \in \mathcal{K}$ with arity $n_{f}$ and $\forall x_{1}, \cdots, x_{n_{f}} \in E$ so that $P(x_{1}), \cdots, P(x_{n_{f}})$ are satisfied, $P(x_{1}, \cdots, x_{n_{f}})$ is also satisfied.

A simple example of structural induction is mathematical induction (raisonnement par récurrence).
Indeed, if you take $\mathcal{B}=\{0\}$ and $\mathcal{K}=\{succ\}$ such as $\forall n \in \mathbb{N}$, $succ(n)=n+1$, then $\mathbb{N}=$<$\mathcal{B},\mathcal{K}$>.
So, mathematical induction is just a proof by induction in $\mathbb{N}$=<${0},{succ}$>.

Exercise: structural induction

Semantics of logic

Boolean algebra

Here is given the truth table of basic operations:

  • $\wedge$ stands for AND (=conjunction)
  • $\vee$ for OR (=disjunction)
  • $\Rightarrow$ for IMPLIES (=implication)
  • $\neg$ for NOT (=negation)

$x$ and $y$ are propositional variables of whom you give values $\in \{0,1\}$. $0$ stands for FALSE and $1$ for TRUE. So, as you can see, the truth table indicates the value of each basic operation according to the values of $x$ and $y$.

$x$ $y$ $x \vee y$ $x \wedge y$ $x \Rightarrow y$ $\neg x$
0 0 0 0 1 1
0 1 1 0 1 1
1 0 1 0 0 0
1 1 1 1 1 0

Propositional calculus

$X=\{x_1, \cdots, x_n\}$ is an infinite set of propositional variables.
The set of formulas $\mathcal{F}$ is inductively defined:

  • Any propositional variable is a formula
  • $\perp$ is a formula
  • If $F$ is a formula, then $\neg F$ too
  • If $F$ and $G$ are formulas, then $F \wedge G$, $F \vee G$, and $F \Rightarrow G$ are also formulas

Remark: Notice that $F \Leftrightarrow G$ has the same signification that ($F \Rightarrow G$) $\wedge$ ($G \Rightarrow F$).

Abusively, an interpretation $I$ is a function that gives the value of a formula. Furthermore, interpretations have some interesting rules:

  • $I(\perp)=0$
  • If $F$ is a formula, then $I(\neg F)=\neg F$
  • If $F$ and $G$ are formulas, then $I(F \wedge G)=I(F) \wedge I(G)$
  • If $F$ and $G$ are formulas, then $I(F \vee G)=I(F) \vee I(G)$
  • If $F$ and $G$ are formulas, then $I(F \Rightarrow G)=I(F) \Rightarrow I(G)$

To be precise, the objective of interpretation is to give meaning to the formal language we've previously defined (the propositional variable and the basic operations also named logical connective).

Here are some vocabulary or other rules :

  • If $I(F)=1$ then $F$ is satisfied by $I$ (I satisfait F), represented by $I \models F$
  • $\Sigma$ a set a formulas, if $\forall F \in \Sigma$, $I(F)=1$ then $\Sigma$ is satisfied by $I$, represented by $I \models \Sigma$
  • If $\forall$ interpretation $I$, $I(F)=1$ then $F$ is a tautology (F est une tautologie)
  • Likewise, if $\forall$ interpretation $I$, $I(F)=0$ then $F$ is a contradiction (F est une contradiction)
  • $\Sigma$ semantically satisfies $F$ (Sigma déduit sémantiquement F), if $\forall$ interpretation $I$ satisfying $\Sigma$, $I$ also satisfies $F$
  • $F$ and $G$ are semantically equivalent (F et G sont semantiquement équivalents), represented by $F \equiv G$, if ${F} \models G$ and ${G} \models F$
  • $\Sigma \models F \Rightarrow G$ if and only if $\Sigma,F \models G$
  • $\Sigma \models F$ if and only if $\Sigma, \neg F$ is a contradiction

Exercise: semantic proof

Further knowledge :

Natural deduction

Your goal is to demonstrate that something is true. You will start from the conclusion=what you think is true and, using Sequent calculus / Séquents prouvables, you will aim to split your conclusion into axioms / axiomes (=something always true).

We are defining F, G, H as propositions and $\Gamma$ a set of propositions. We will use the notation


Everything at the left of $\vdash$ is true, at first, you got nothing here, and your expression at the right of $\vdash$, but your goal will be to move as many propositions as possible inside Gamma using the sequent calculus then when you can, you need to try to get axioms (the sequent ax) for each expression with a $\vdash$.

@1.\quad\frac{}{\Gamma, F \vdash F} (ax)@ @2.\quad\frac{\Gamma, F \vdash F}{\Gamma, G \vdash F} (aff)@ @3.\quad\frac{\Gamma, F \vdash G}{\Gamma \vdash F \Rightarrow G} (\Rightarrow_i)@
@4.\quad\frac{\Gamma \vdash F \Rightarrow G \quad \Gamma \vdash F}{\Gamma \vdash G} (\Rightarrow_e)@ @5.\quad\frac{\Gamma \vdash F \quad \Gamma \vdash G}{\Gamma \vdash F \wedge G} (\wedge_i)@ @6.\quad\frac{\Gamma \vdash F \wedge G}{\Gamma \vdash F} (\wedge_e^l)@
@7.\quad\frac{\Gamma \vdash F \wedge G}{\Gamma \vdash G} (\wedge_e^r)@ @8.\quad\frac{\Gamma \vdash F}{\Gamma \vdash F \vee G} (\vee_i^l)@ @9.\quad\frac{\Gamma \vdash G}{\Gamma \vdash F \vee G} (\vee_i^r)@
@10.\quad\frac{\Gamma \vdash F \vee G \quad \Gamma, F \vdash H \quad \Gamma, G \vdash H}{\Gamma \vdash H} (\vee_e)@ @11.\quad\frac{\Gamma, F \vdash \perp}{\Gamma \vdash \neg F} (\neg_i)@ @12.\quad\frac{\Gamma \vdash \neg F \quad \Gamma \vdash F}{\Gamma \vdash \perp} (\neg_e)@
@13.\quad\frac{\Gamma, \neg F \vdash \perp}{\Gamma \vdash F} (\perp_c)@

You must write the letters at the right of a sequent each time you are using it. Just so you know

  • i means introduction (of a new variable)
  • e means Elimination (of a variable)
  • l means left
  • r means right

First-order logic



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