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tokenize "derivation"
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content/first-order-logic/axiomatic-deduction/axioms-rules-propositional.tex

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\end{defn}
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\begin{defn}[Modus ponens]
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If $!B$ and $!B \lif !A$ already occur in a derivation, then $!A$ is
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If $!B$ and $!B \lif !A$ already occur in !!a{derivation}, then $!A$ is
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a correct inference step.
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\end{defn}
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content/first-order-logic/axiomatic-deduction/deduction-theorem.tex

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induction basis applies.) Then some previous steps in the
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!!{derivation} are $!C \lif !B$ and $!C$, for some !!{formula}~$!C$,
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i.e., $\Gamma \cup \{!A\} \Proves !C \lif !B$ and $\Gamma \cup \{!A\}
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\Proves !C$, and the respective derivations are shorter, so the
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\Proves !C$, and the respective !!{derivation}s are shorter, so the
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inductive hypothesis applies to them. We thus have both:
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\begin{align*}
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& \Gamma \Proves !A \lif (!C \lif !B); \\

content/first-order-logic/axiomatic-deduction/proof-theoretic-notions.tex

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\begin{proof}
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Suppose $\{!A\} \cup \Delta \Proves !B$. Then there is
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!!a{derivation} $!B_1$, \dots, $!B_l = !B$ from~$\{!A\} \cup
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\Delta$. Some of the steps in that derivation will be correct
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\Delta$. Some of the steps in that !!{derivation} will be correct
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because of a rule which refers to a prior line~$!B_i = !A$. By
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hypothesis, there is !!a{derivation} of~$!A$ from~$\Gamma$, i.e.,
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!!a{derivation}~$!A_1$, \dots, $!A_k = !A$ where every $!A_i$ is an

content/first-order-logic/axiomatic-deduction/proving-things-quant.tex

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\olsection{\usetoken{P}{derivation} with Quantifiers}
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\begin{ex}
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Let us give a derivation of $(\lforall[x][!A(x)] \land
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Let us give !!a{derivation} of $(\lforall[x][!A(x)] \land
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\lforall[y][!B(y)]) \lif \lforall[x][(!A(x) \land !B(x))]$.
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First, note that

content/first-order-logic/axiomatic-deduction/proving-things.tex

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Why? Two applications of MP yield the last part, which is what we
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want. And we easily see that $\lnot !D \lif (!D \lif !E)$ is an
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instance of \olref[prp]{ax:lnot2}, and $!E \lif (!D \lif !E)$ is an
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instance of \olref[prp]{ax:lif1}. So our derivation is:
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instance of \olref[prp]{ax:lif1}. So our !!{derivation} is:
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\begin{derivation}
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1. & $\lnot !D \lif (!D \lif !E)$ & \olref[prp]{ax:lnot2} \\
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2. & $(\lnot !D \lif (!D \lif !E)) \lif {}$\\
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also !!{derivable}? Well, the first of these is already an instance of
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\olref[prp]{ax:lif1}, whatever we decide $!B$ to be. And $!D \lif !B$ would
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be another instance of \olref[prp]{ax:lif1} if $!B$ were $(!D \lif !D)$.
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So, our derivation is:
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So, our !!{derivation} is:
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\begin{derivation}
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1. & $!D \lif ((!D \lif !D) \lif !D)$ & \olref[prp]{ax:lif1}\\
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2. & $(!D \lif ((!D \lif !D) \lif !D)) \lif {}$\\
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\end{ex}
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\begin{ex}\ollabel{ex:chain}
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Sometimes we want to show that there is a derivation of some
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Sometimes we want to show that there is !!a{derivation} of some
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!!{formula} from some other !!{formula}s~$\Gamma$. For instance, let's
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show that we can !!{derive} $!A \lif !C$ from $\Gamma = \{!A \lif !B,
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!B \lif !C\}$.

content/first-order-logic/axiomatic-deduction/soundness.tex

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\begin{proof}
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By induction on the length of the !!{derivation} of $!A$ from
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$\Gamma$. If there are no steps justified by inferences, then all
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!!{formula}s in the derivation are either instances of axioms or are
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!!{formula}s in the !!{derivation} are either instances of axioms or are
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in~$\Gamma$. By the previous proposition, all the axioms are
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\iftag{FOL}{valid}{tautologies}, and hence if $!A$ is an axiom then
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$\Gamma \Entails !A$. If $!A \in \Gamma$, then trivially $\Gamma
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\Entails !A$.
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If the last step of the derivation of~$!A$ is justified by modus
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If the last step of the !!{derivation} of~$!A$ is justified by modus
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ponens, then there are !!{formula}s $!B$ and $!B \lif !A$ in the
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!!{derivation}, and the induction hypothesis applies to the part of
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the !!{derivation} ending in those !!{formula}s (since they contain at

content/first-order-logic/natural-deduction/derivations.tex

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\RightLabel{\Intro{\land}}
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\BinaryInfC{$!D \land !C$}
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\end{prooftree}
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is also a correct derivation.
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is also a correct !!{derivation}.
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We can now apply another rule, say, $\Intro{\lif}$, which allows us to
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conclude a conditional and allows us to !!{discharge} any assumption
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Remember that discharging of assumptions is a permission, not a
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requirement: we don't have to discharge the assumptions. In
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particular, we can apply a rule even if the assumptions are not
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present in the derivation. For instance, the following is legal, even
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present in the !!{derivation}. For instance, the following is legal, even
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though there is no assumption~$!A$ to be !!{discharged}:
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\begin{prooftree}
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\AxiomC{$!B$}

content/first-order-logic/natural-deduction/identity.tex

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\end{prooftree}
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The sub-!!{derivation} on the top right is completed by using its
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assumptions to show that $\eq[a][c]$ and $\eq[b][c]$. This requires two
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separate !!{derivation}s. The derivation for $\eq[a][c]$ is as
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separate !!{derivation}s. The !!{derivation} for $\eq[a][c]$ is as
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follows:
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\begin{prooftree}
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\AxiomC{$\Discharge{\lforall[y][(!A(y) \lif \eq[y][c]])}{2}$}

content/first-order-logic/natural-deduction/provability-consistency.tex

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\end{prooftree}
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Now assume $\Gamma \cup \{\lnot !A\}$ is inconsistent, and let
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$\delta_1$ be the corresponding derivation of $\lfalse$ from
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$\delta_1$ be the corresponding !!{derivation} of~$\lfalse$ from
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!!{undischarged} assumptions in~$\Gamma \cup \{\lnot !A\}$. We obtain
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!!a{derivation} of~$!A$ from~$\Gamma$ alone by using~$\FalseCl$:
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\begin{prooftree}

content/first-order-logic/natural-deduction/proving-things-quant.tex

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violated. Since the only rule we applied that is subject to the
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eigenvariable condition was \Elim{\exists}, and the eigenvariable~$a$
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does not occur in any assumptions it depends on, this is a
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correct derivation.
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correct !!{derivation}.
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\end{ex}
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\begin{ex}
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\end{prooftree}
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Since we ensured at each step that the eigenvariable
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conditions were not violated, we can be confident that this
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is a correct derivation.
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is a correct !!{derivation}.
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\end{ex}
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\begin{ex}

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