Math 323 - Arizona Math
... 13. a. To prove that x is an element of the interval (3, ∞), one must prove both x > 3 and also x < ∞. b. If x is an element of the interval (3, ∞), then x > 3 and x “goes on forever ”. 14. a. The equation x2 - 4 x = y has no solutions because there are two variables and only one equation. b. One ca ...
... 13. a. To prove that x is an element of the interval (3, ∞), one must prove both x > 3 and also x < ∞. b. If x is an element of the interval (3, ∞), then x > 3 and x “goes on forever ”. 14. a. The equation x2 - 4 x = y has no solutions because there are two variables and only one equation. b. One ca ...
On Perfect Introspection with Quantifying-in
... of an agent can be reduced to the objective ones. In ILL88], it was shown that this result holds even if we weaken the logic in the sense that beliefs are not necessarily closed under (classical) logical consequence. In the first-order case, however, we run into problems because of the phenomenon of ...
... of an agent can be reduced to the objective ones. In ILL88], it was shown that this result holds even if we weaken the logic in the sense that beliefs are not necessarily closed under (classical) logical consequence. In the first-order case, however, we run into problems because of the phenomenon of ...
Plural Quantifiers
... This will be false in standard models of arithmetic. Consider any group of standard numbers. If it contains 0, then it can’t be a group of numbers that only doodle each other, since 0 doodles itself. If it doesn’t contain 0, then let k be the least number it contains. Since k is the least number in ...
... This will be false in standard models of arithmetic. Consider any group of standard numbers. If it contains 0, then it can’t be a group of numbers that only doodle each other, since 0 doodles itself. If it doesn’t contain 0, then let k be the least number it contains. Since k is the least number in ...
4. Homework Assignment #4 Math 4/515 Problem 4.1. If x > 0 is
... countable cover of [0, 1]. Extract a finite sub cover. Thus for some n, we have the inclusion [0, 1] ⊂ R1 ∪ · · · ∪ Rk ∪ I1 ∪ · · · ∪ In . Try to show that this can not be true by integrating the inequality: χ[0,1] ≤ χR1 + · · · + χRk + χI1 + · · · + χIn . (3) Prove or disprove the following asserti ...
... countable cover of [0, 1]. Extract a finite sub cover. Thus for some n, we have the inclusion [0, 1] ⊂ R1 ∪ · · · ∪ Rk ∪ I1 ∪ · · · ∪ In . Try to show that this can not be true by integrating the inequality: χ[0,1] ≤ χR1 + · · · + χRk + χI1 + · · · + χIn . (3) Prove or disprove the following asserti ...
Exercises: Sufficiently expressive/strong
... (b) Is your favourite proof system for classical propositional logic effectively decidable? (c) Suppose Q is a finitely axiomatizable theory with a standard first-order logic; then show that there is a single sentence Q̂ such that Q ` ϕ if and only if ` Q̂ → ϕ (where ` is deducibility in your favour ...
... (b) Is your favourite proof system for classical propositional logic effectively decidable? (c) Suppose Q is a finitely axiomatizable theory with a standard first-order logic; then show that there is a single sentence Q̂ such that Q ` ϕ if and only if ` Q̂ → ϕ (where ` is deducibility in your favour ...
Answer Sets for Propositional Theories
... Π, Y |= Π X iff X |= Π and Y |= Π X . This corollary suggests another way to verify that the definition of an answer set proposed in this paper is equivalent to the usual one in the case of programs with nested expressions. If X 6|= Π then X is not an answer set for Π under either semantics. Otherwi ...
... Π, Y |= Π X iff X |= Π and Y |= Π X . This corollary suggests another way to verify that the definition of an answer set proposed in this paper is equivalent to the usual one in the case of programs with nested expressions. If X 6|= Π then X is not an answer set for Π under either semantics. Otherwi ...
