2020-02-02
∃x,y(x∈A ^ y∈A ^ ￢(x=y) ^ A⊆{x, y})
∃x,y(x∈A ^ y∈A ^ ￢(x=y) ^ A⊆{x, y})
∀x, y, z (x ∈ A ^ y ∈ A ^ x ≠ y ⇒ z = x ∨ z = y)
∀x, y, z (x ∈ A ^ y ∈ A ^ x ≠ y ⇒ z = x ∨ z = y)
∃ x,y (x ∈ A ∧ y ∈ A ∧ x ¬= y ∧ (∀z z ∈ A => z=x ∧ z=y))
∃ x,y (x ∈ A ∧ y ∈ A ∧ x ¬= y ∧ (∀z z ∈ A => z=x ∧ z=y))
∃x,y(x!=y ∧ x∈A ∧ y∈A ∧ A={x,y})
(Edited: 2020-02-03) ∃x,y(x!=y ∧ x∈A ∧ y∈A ∧ A={x,y})
In classwork.
(a ≠b )∈A
A={a,b}
In classwork.
(a ≠b )∈A
A={a,b}
∃x, y ( x∈A ∧ y∈A ∧ ¬(x = y) ∧ ( ∀z ( z∈A ⇒ z=x ∨ z=y) ) )
∃x, y ( x∈A ∧ y∈A ∧ ¬(x = y) ∧ ( ∀z ( z∈A ⇒ z=x ∨ z=y) ) )
I was thinking that if x and y do not equal to each other, then they are different elements. Then if x and y are elements of A, then A will at least have 2 elements. :
∃X,Y (Y ∈ A ∧X ∈ A.X ≠ Y, (∀z z = X ∨ z = Y))
I was thinking that if x and y do not equal to each other, then they are different elements. Then if x and y are elements of A, then A will at least have 2 elements. :
∃X,Y (Y ∈ A ∧X ∈ A.
X ≠ Y,
(∀z z = X ∨ z = Y))
For every set, there exists one set A in which A equals {a,b} and only {a,b}
Seems to be totally wrong but this is what I was thinking at the time.
For every set, there exists one set A in which A equals {a,b} and only {a,b}
Seems to be totally wrong but this is what I was thinking at the time.
((resource:A1.png|Resource Description for A1.png))
((resource:83881654_1859875540804441_4504783801393610752_n.jpg|Resource Description for 83881654_1859875540804441_4504783801393610752_n.jpg))
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