Inference Rules for Quantified Propositions

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”.

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).

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)

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).

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).

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

______________________ 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.

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

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.]

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).

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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