Feb:13(Fri)
{ x : x is a binary digit }
{ x : x is a letter in your last name }
{ x : x ∈ N and x² < 20 }
{ x : x = 2 or x = 7 }
{ x : x = 2 and x = 7 }
{ x + y : x = 2 or x = 7 and y = 2 or y = 7 }
{ S : S is the set of instructors of MTH 225 this semester
or S is the set of instructors of CS 163 this semester }
or S is the set of instructors of CS 361 this semester }
{ S : S is a set containing two letters of "helo" }
⊂", "⊆",
or "⊈" (a.k.a. "⊄")
in each box in (a reproduction of) the following 'table':
+---+ +---+ +---+ +---+ A | | A A | | B A | | C A | | D +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ B | | A B | | B B | | C B | | D +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ C | | A C | | B C | | C C | | D +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ D | | A D | | B D | | C D | | D +---+ +---+ +---+ +---+
|∅|
| { ∅ } |
| { ∅, {∅} } |
| { ∅, {∅}, {∅, {∅}} } |
{1,2}, B := {a,b,c},
C := {α,β},
and D := {$}.
Give list representations
of each of the following sets.
Be careful to use correct notation
(appropriate parentheses, braces).
∅"
instead of "{ }"
for any set that is empty.
∅"
instead of "{ }"
for any set that is empty.
∀S1∀S2[S1 ⊆ S2 ↔ ∀x(x ∈ S1 → x ∈ S2)] ∀S1∀S2[S1 = S2 ↔ ∀x(x ∈ S1 ↔ x ∈ S2)] ∀S1∀S2[S1 = S2 ↔ (S1 ⊆ S2 ∧ S2 ⊆ S1)] ∀S1∀S2∀x[x ∈ S1 ∪ S2 ↔ (x ∈ S1 ∨ x ∈ S2)] ∀S1∀S2∀x[x ∈ S1 ∩ S2 ↔ (x ∈ S1 ∧ x ∈ S2)] ∀S1∀S2∀x[x ∈ S1 - S2 ↔ (x ∈ S1 ∧ x ∉ S2)] ∀S∀x[x ∈ ~S ↔ x ∉ S] ∀S1∀S2[~(S1 ∩ S2) = ~S1 ∪ ~S2] ∀S1∀S2[S1 - S2 = S1 ∩ ~S2] ∀S1∀S2∀S3[S1 ∩ (S2 ∪ S3) = (S1 ∩ S2) ∪ (S1 ∩ S3)]Using ProofBuilder, produce proofs of the formulas below. Use the axioms and identities above — and further, in your proofs of the numbered formulas below, you are allowed to use lower-numbered formulas. Thus, the proof of Example 14 of page 126 used 'earlier' formulas, not just axioms.
Rem. you can copy material from here and lecture notes and paste such into ProofBuilder.
∀S[∅ ⊆ S]
∀S[S ⊆ S]
A ⊆ B ∧ B ⊆ C → A ⊆ C
(2.1.15)
∀S[S ∪ S = S]
∀S[∅ ∪ S = S]
∀S[S ∩ ∅ = ∅]
∀S1∀S2[S1 ∪ (S1 ∩ S2) = S1]
∀S[~~S = S]
∀S[S ∩ ~S = ∅]
∀S1∀S2[S1 ∩ (S2 - S1) = ∅]
A - (A - B) = A ∩ B