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,円P(x)} |
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,円P(x)} can be derived.
Generalization with hypotheses
[edit ]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,円\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:
- {\displaystyle \exists z,円\exists w,円(z\not =w)} (Hypothesis)
- {\displaystyle \exists w,円(y\not =w)} (Existential instantiation)
- {\displaystyle y\not =x} (Existential instantiation)
- {\displaystyle \forall x,円(x\not =x)} (Faulty universal generalization)
This purports to show that {\displaystyle \exists z,円\exists w,円(z\not =w)\vdash \forall x,円(x\not =x),} which is an unsound deduction. Note that {\displaystyle \Gamma \vdash \forall y,円\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).
Example of a proof
[edit ]Prove: {\displaystyle \forall x,円(P(x)\rightarrow Q(x))\rightarrow (\forall x,円P(x)\rightarrow \forall x,円Q(x))} is derivable from {\displaystyle \forall x,円(P(x)\rightarrow Q(x))} and {\displaystyle \forall x,円P(x)}.
Proof:
| Step | Formula | Justification |
|---|---|---|
| 1 | {\displaystyle \forall x,円(P(x)\rightarrow Q(x))} | Hypothesis |
| 2 | {\displaystyle \forall x,円P(x)} | Hypothesis |
| 3 | {\displaystyle (\forall x,円(P(x)\rightarrow Q(x)))\rightarrow (P(y)\rightarrow Q(y))} | From (1) by Universal instantiation |
| 4 | {\displaystyle P(y)\rightarrow Q(y)} | From (1) and (3) by Modus ponens |
| 5 | {\displaystyle (\forall x,円P(x))\rightarrow P(y)} | From (2) by 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,円Q(x)} | From (7) by Generalization |
| 9 | {\displaystyle \forall x,円(P(x)\rightarrow Q(x)),\forall x,円P(x)\vdash \forall x,円Q(x)} | Summary of (1) through (8) |
| 10 | {\displaystyle \forall x,円(P(x)\rightarrow Q(x))\vdash \forall x,円P(x)\rightarrow \forall x,円Q(x)} | From (9) by Deduction theorem |
| 11 | {\displaystyle \vdash \forall x,円(P(x)\rightarrow Q(x))\rightarrow (\forall x,円P(x)\rightarrow \forall x,円Q(x))} | From (10) by Deduction theorem |
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.