Article TARSKI, MML version 4.99.1005
:: TARSKI:th 2
theorem
for b1, b2 being set
st for b3 being set holds
b3 in b1
iff
b3 in b2
holds b1 = b2;
:: TARSKI:funcnot 1 => TARSKI:func 1
definition
let a1 be set;
func {A1} -> set means
for b1 being set holds
b1 in it
iff
b1 = a1;
end;
:: TARSKI:def 1
theorem
for b1, b2 being set holds
b2 = {b1}
iff
for b3 being set holds
b3 in b2
iff
b3 = b1;
:: TARSKI:funcnot 2 => TARSKI:func 2
definition
let a1, a2 be set;
func {A1,A2} -> set means
for b1 being set holds
b1 in it
iff
(b1 = a1 or b1 = a2);
commutativity;
:: for a1, a2 being set holds
:: {a1,a2} = {a2,a1};
end;
:: TARSKI:def 2
theorem
for b1, b2, b3 being set holds
b3 = {b1,b2}
iff
for b4 being set holds
b4 in b3
iff
(b4 = b1 or b4 = b2);
:: TARSKI:prednot 1 => TARSKI:pred 1
definition
let a1, a2 be set;
pred A1 c= A2 means
for b1 being set
st b1 in a1
holds b1 in a2;
reflexivity;
:: for a1 being set holds
:: a1 c= a1;
end;
:: TARSKI:dfs 3
definiens
let a1, a2 be set;
To prove
a1 c= a2
it is sufficient to prove
thus for b1 being set
st b1 in a1
holds b1 in a2;
:: TARSKI:def 3
theorem
for b1, b2 being set holds
b1 c= b2
iff
for b3 being set
st b3 in b1
holds b3 in b2;
:: TARSKI:funcnot 3 => TARSKI:func 3
definition
let a1 be set;
func union A1 -> set means
for b1 being set holds
b1 in it
iff
ex b2 being set st
b1 in b2 & b2 in a1;
end;
:: TARSKI:def 4
theorem
for b1, b2 being set holds
b2 = union b1
iff
for b3 being set holds
b3 in b2
iff
ex b4 being set st
b3 in b4 & b4 in b1;
:: TARSKI:th 7
theorem
for b1, b2 being set
st b1 in b2
holds ex b3 being set st
b3 in b2 &
(for b4 being set
st b4 in b2
holds not b4 in b3);
:: TARSKI:sch 1
scheme TARSKI:sch 1
{F1 -> set}:
ex b1 being set st
for b2 being set holds
b2 in b1
iff
ex b3 being set st
b3 in F1() & P1[b3, b2]
provided
for b1, b2, b3 being set
st P1[b1, b2] & P1[b1, b3]
holds b2 = b3;
:: TARSKI:funcnot 4 => TARSKI:func 4
definition
let a1, a2 be set;
func [A1,A2] -> set equals
{{a1,a2},{a1}};
end;
:: TARSKI:def 5
theorem
for b1, b2 being set holds
[b1,b2] = {{b1,b2},{b1}};
:: TARSKI:prednot 2 => TARSKI:pred 2
definition
let a1, a2 be set;
pred A1,A2 are_equipotent means
ex b1 being set st
(for b2 being set
st b2 in a1
holds ex b3 being set st
b3 in a2 & [b2,b3] in b1) &
(for b2 being set
st b2 in a2
holds ex b3 being set st
b3 in a1 & [b3,b2] in b1) &
(for b2, b3, b4, b5 being set
st [b2,b3] in b1 & [b4,b5] in b1
holds b2 = b4
iff
b3 = b5);
end;
:: TARSKI:dfs 6
definiens
let a1, a2 be set;
To prove
a1,a2 are_equipotent
it is sufficient to prove
thus ex b1 being set st
(for b2 being set
st b2 in a1
holds ex b3 being set st
b3 in a2 & [b2,b3] in b1) &
(for b2 being set
st b2 in a2
holds ex b3 being set st
b3 in a1 & [b3,b2] in b1) &
(for b2, b3, b4, b5 being set
st [b2,b3] in b1 & [b4,b5] in b1
holds b2 = b4
iff
b3 = b5);
:: TARSKI:def 6
theorem
for b1, b2 being set holds
b1,b2 are_equipotent
iff
ex b3 being set st
(for b4 being set
st b4 in b1
holds ex b5 being set st
b5 in b2 & [b4,b5] in b3) &
(for b4 being set
st b4 in b2
holds ex b5 being set st
b5 in b1 & [b5,b4] in b3) &
(for b4, b5, b6, b7 being set
st [b4,b5] in b3 & [b6,b7] in b3
holds b4 = b6
iff
b5 = b7);
:: TARSKI:th 9
theorem
for b1 being set holds
ex b2 being set st
b1 in b2 &
(for b3, b4 being set
st b3 in b2 & b4 c= b3
holds b4 in b2) &
(for b3 being set
st b3 in b2
holds ex b4 being set st
b4 in b2 &
(for b5 being set
st b5 c= b3
holds b5 in b4)) &
(for b3 being set
st b3 c= b2 & not b3,b2 are_equipotent
holds b3 in b2);