In predicate logic, generalization (also universal generalization, universal introduction,^[1][2][3] GEN, UG) is a valid inference rule. It states that if ${\displaystyle \vdash \!P(x)}$ has been derived, then ${\displaystyle \vdash \!\forall x\,P(x)}$ can be derived.

This article includes a list of general references, but it lacks sufficient corresponding inline citations. (March 2023)

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Universal generalization

Type	Rule of inference
Field	Predicate logic
Statement	Suppose ${\displaystyle P}$ is true of any arbitrarily selected ${\displaystyle p}$, then ${\displaystyle P}$ is true of everything.
Symbolic statement	${\displaystyle \vdash \!P(x)}$, ${\displaystyle \vdash \!\forall x\,P(x)}$

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The full generalization rule allows for hypotheses to the left of the turnstile, but with restrictions. Assume ${\displaystyle \Gamma }$ is a set of formulas, ${\displaystyle \varphi }$ a formula, and ${\displaystyle \Gamma \vdash \varphi (y)}$ has been derived. The generalization rule states that ${\displaystyle \Gamma \vdash \forall x\,\varphi (x)}$ can be derived if ${\displaystyle y}$ is not mentioned in ${\displaystyle \Gamma }$ and ${\displaystyle x}$ does not occur in ${\displaystyle \varphi }$.

These restrictions are necessary for soundness. Without the first restriction, one could conclude ${\displaystyle \forall xP(x)}$ from the hypothesis ${\displaystyle P(y)}$. Without the second restriction, one could make the following deduction:

1. ${\displaystyle \exists z\,\exists w\,(z\not =w)}$ (Hypothesis)
2. ${\displaystyle \exists w\,(y\not =w)}$ (Existential instantiation)
3. ${\displaystyle y\not =x}$ (Existential instantiation)
4. ${\displaystyle \forall x\,(x\not =x)}$ (Faulty universal generalization)

This purports to show that ${\displaystyle \exists z\,\exists w\,(z\not =w)\vdash \forall x\,(x\not =x),}$ which is an unsound deduction. Note that ${\displaystyle \Gamma \vdash \forall y\,\varphi (y)}$ is permissible if ${\displaystyle y}$ is not mentioned in ${\displaystyle \Gamma }$ (the second restriction need not apply, as the semantic structure of ${\displaystyle \varphi (y)}$ is not being changed by the substitution of any variables).

Prove: ${\displaystyle \forall x\,(P(x)\rightarrow Q(x))\rightarrow (\forall x\,P(x)\rightarrow \forall x\,Q(x))}$ is derivable from ${\displaystyle \forall x\,(P(x)\rightarrow Q(x))}$ and ${\displaystyle \forall x\,P(x)}$.

Proof:

More information , ...

Step	Formula	Justification
1	${\displaystyle \forall x\,(P(x)\rightarrow Q(x))}$	Hypothesis
2	${\displaystyle \forall x\,P(x)}$	Hypothesis
3	${\displaystyle (\forall x\,(P(x)\rightarrow Q(x)))\rightarrow (P(y)\rightarrow Q(y))}$	Universal instantiation
4	${\displaystyle P(y)\rightarrow Q(y)}$	From (1) and (3) by Modus ponens
5	${\displaystyle (\forall x\,P(x))\rightarrow P(y)}$	Universal instantiation
6	${\displaystyle P(y)\ }$	From (2) and (5) by Modus ponens
7	${\displaystyle Q(y)\ }$	From (6) and (4) by Modus ponens
8	${\displaystyle \forall x\,Q(x)}$	From (7) by Generalization
9	${\displaystyle \forall x\,(P(x)\rightarrow Q(x)),\forall x\,P(x)\vdash \forall x\,Q(x)}$	Summary of (1) through (8)
10	${\displaystyle \forall x\,(P(x)\rightarrow Q(x))\vdash \forall x\,P(x)\rightarrow \forall x\,Q(x)}$	From (9) by Deduction theorem
11	${\displaystyle \vdash \forall x\,(P(x)\rightarrow Q(x))\rightarrow (\forall x\,P(x)\rightarrow \forall x\,Q(x))}$	From (10) by Deduction theorem

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In this proof, universal generalization was used in step 8. The deduction theorem was applicable in steps 10 and 11 because the formulas being moved have no free variables.

• First-order logic
• Hasty generalization
• Universal instantiation