1. Inference Rules for Quantified Propositions

2. Universal Specification
If a statement of the form x, P(x) is true, then P(c) is true for arbitrary c in the universe of discourse.
This can be written:
x, P(x)
________
 P(c) for all c.
Example: M(x): x is mortal. From x, M(x), we infer that “Socrates is mortal”.

3. Universal Generalization
If a statement P(c) is true for each element c of the universe, then x, P(x).
This can be written:
P(c) for all c
________
 x, P(x).

4. Example of Universal Generalization
Prove: if R is an antisymmetric relation, so is R-1.
Let xRy be an arbitrary element of R where x  y.
yR-1x. (Defn. of inverse relation.)
(y,x)  R. (R is antisymmetric)
(x,y)  R-1. (Step 3 & defn. of inverse relation.)
(x,y), (xR-1y  yR-1x)  x = y. (UG)
R-1 is antisymmetric. (Step 5 & defn. of antisymmetric)

5. Existential Specification
If x, P(x) is true then there is an element c such that P(c) is true.
This can be written:
x, P(x)
________
 P(c), for some c.
Element c is not arbitrary.
We know only that some c satisfies P.
We do not necessarily know which one (e.g., from a non-constructive proof).

6. Existential Generalization
If P(c) is true for some c, then x, P(x).
This can be written:
P(c), for some c
________
 x, P(x).

7. Protocol To infer from quantified premises:
Properly remove quantifiers.
Use existential or universal specification
Argue with the resulting propositions.
Properly prefix the correct quantifiers.
Use existential & universal generalization

8. ______________________
Example arguments
All humans are fallible.
All government agents are human.
Therefore, all government agents are fallible.
______________________
H(x): x is a human.
F(x): x is fallible.
G(x): x is a government agent.

9. x, ( H(x)  F(x) ).
x, ( G(x)  H(x) ).
_________________
x, ( G(x)  F(x) ).

10. Proof: 1. x, ( H(x)  F(x) ) [premise 1]
2. H(c)  F(c) [ step 1 & U.S.]
3. x, ( G(x)  H(x) ). [ premise 2]
4. G(c)  H(c) [ step 3 & U.S.]
5. G(c)  F(c) [steps 2, 4, & transitivity]
6. x, ( G(x)  F(x) ). [step 5 & U.G.]

11. A more compact representation:
In English:
All CCS classes are easy.
This is a CCS class.
Therefore, this class is easy.
A more compact representation:
x, C(x)  E(x).
C(CCS CS 2).
Therefore, E(CCS CS 2).

12. Proof 1. x, C(x)  E(x) [premise 1]
2. C(CCS CS 2)  E(CCS CS 2) [step 1, U.S.]
3. C(CCS CS 2) [premise 2]
4. E(CCS CS 2) [step 2, 3, & modus ponens]

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