Article ROUGHS_1, MML version 4.99.1005
:: ROUGHS_1:funcreg 1
registration
let a1 be set;
cluster RelStr(#a1,id a1#) -> strict discrete;
end;
:: ROUGHS_1:th 1
theorem
for b1 being set
st nabla b1 c= id b1
holds b1 is trivial;
:: ROUGHS_1:attrnot 1 => ROUGHS_1:attr 1
definition
let a1 be RelStr;
attr a1 is diagonal means
the InternalRel of a1 c= id the carrier of a1;
end;
:: ROUGHS_1:dfs 1
definiens
let a1 be RelStr;
To prove
a1 is diagonal
it is sufficient to prove
thus the InternalRel of a1 c= id the carrier of a1;
:: ROUGHS_1:def 1
theorem
for b1 being RelStr holds
b1 is diagonal
iff
the InternalRel of b1 c= id the carrier of b1;
:: ROUGHS_1:funcreg 2
registration
let a1 be non trivial set;
cluster RelStr(#a1,nabla a1#) -> strict non diagonal;
end;
:: ROUGHS_1:th 2
theorem
for b1 being reflexive RelStr holds
id the carrier of b1 c= the InternalRel of b1;
:: ROUGHS_1:condreg 1
registration
cluster reflexive non discrete -> non trivial (RelStr);
end;
:: ROUGHS_1:condreg 2
registration
cluster trivial reflexive -> discrete (RelStr);
end;
:: ROUGHS_1:th 3
theorem
for b1 being set
for b2 being reflexive total Relation of b1,b1 holds
id b1 c= b2;
:: ROUGHS_1:condreg 3
registration
cluster discrete -> diagonal (RelStr);
end;
:: ROUGHS_1:condreg 4
registration
cluster non diagonal -> non discrete (RelStr);
end;
:: ROUGHS_1:exreg 1
registration
cluster non empty non diagonal RelStr;
end;
:: ROUGHS_1:th 4
theorem
for b1 being non empty non diagonal RelStr holds
ex b2, b3 being Element of the carrier of b1 st
b2 <> b3 & [b2,b3] in the InternalRel of b1;
:: ROUGHS_1:th 5
theorem
for b1 being set
for b2, b3 being FinSequence of b1 holds
Union (b2 ^ b3) = (Union b2) \/ Union b3;
:: ROUGHS_1:th 6
theorem
for b1, b2 being Relation-like Function-like set
st b2 is disjoint_valued & b1 c= b2
holds b1 is disjoint_valued;
:: ROUGHS_1:condreg 5
registration
cluster Relation-like Function-like empty -> disjoint_valued (set);
end;
:: ROUGHS_1:exreg 2
registration
let a1 be set;
cluster Relation-like Function-like finite disjoint_valued FinSequence-like FinSequence of a1;
end;
:: ROUGHS_1:exreg 3
registration
let a1 be non empty set;
cluster Relation-like Function-like non empty finite disjoint_valued FinSequence-like FinSequence of a1;
end;
:: ROUGHS_1:funcnot 1 => ROUGHS_1:func 1
definition
let a1 be set;
let a2 be FinSequence of bool a1;
let a3 be natural set;
redefine func a2 . a3 -> Element of bool a1;
end;
:: ROUGHS_1:funcnot 2 => ROUGHS_1:func 2
definition
let a1 be set;
let a2 be FinSequence of bool a1;
redefine func Union a2 -> Element of bool a1;
end;
:: ROUGHS_1:funcreg 3
registration
let a1 be finite set;
let a2 be Relation of a1,a1;
cluster RelStr(#a1,a2#) -> finite strict;
end;
:: ROUGHS_1:th 7
theorem
for b1, b2, b3 being set
for b4 being reflexive symmetric total Relation of b1,b1
st b2 in Class(b4,b3)
holds b3 in Class(b4,b2);
:: ROUGHS_1:attrnot 2 => ROUGHS_1:attr 2
definition
let a1 be RelStr;
attr a1 is with_equivalence means
the InternalRel of a1 is symmetric transitive total Relation of the carrier of a1,the carrier of a1;
end;
:: ROUGHS_1:dfs 2
definiens
let a1 be RelStr;
To prove
a1 is with_equivalence
it is sufficient to prove
thus the InternalRel of a1 is symmetric transitive total Relation of the carrier of a1,the carrier of a1;
:: ROUGHS_1:def 2
theorem
for b1 being RelStr holds
b1 is with_equivalence
iff
the InternalRel of b1 is symmetric transitive total Relation of the carrier of b1,the carrier of b1;
:: ROUGHS_1:attrnot 3 => ROUGHS_1:attr 3
definition
let a1 be RelStr;
attr a1 is with_tolerance means
the InternalRel of a1 is reflexive symmetric total Relation of the carrier of a1,the carrier of a1;
end;
:: ROUGHS_1:dfs 3
definiens
let a1 be RelStr;
To prove
a1 is with_tolerance
it is sufficient to prove
thus the InternalRel of a1 is reflexive symmetric total Relation of the carrier of a1,the carrier of a1;
:: ROUGHS_1:def 3
theorem
for b1 being RelStr holds
b1 is with_tolerance
iff
the InternalRel of b1 is reflexive symmetric total Relation of the carrier of b1,the carrier of b1;
:: ROUGHS_1:condreg 6
registration
cluster with_equivalence -> with_tolerance (RelStr);
end;
:: ROUGHS_1:funcreg 4
registration
let a1 be set;
cluster RelStr(#a1,id a1#) -> strict with_equivalence;
end;
:: ROUGHS_1:exreg 4
registration
cluster non empty finite discrete with_equivalence RelStr;
end;
:: ROUGHS_1:exreg 5
registration
cluster non empty finite non diagonal with_equivalence RelStr;
end;
:: ROUGHS_1:modenot 1
definition
mode Approximation_Space is non empty with_equivalence RelStr;
end;
:: ROUGHS_1:modenot 2
definition
mode Tolerance_Space is non empty with_tolerance RelStr;
end;
:: ROUGHS_1:funcreg 5
registration
let a1 be non empty with_tolerance RelStr;
cluster the InternalRel of a1 -> reflexive symmetric total;
end;
:: ROUGHS_1:funcreg 6
registration
let a1 be non empty with_equivalence RelStr;
cluster the InternalRel of a1 -> transitive;
end;
:: ROUGHS_1:funcnot 3 => ROUGHS_1:func 3
definition
let a1 be non empty RelStr;
let a2 be Element of bool the carrier of a1;
func LAp A2 -> Element of bool the carrier of a1 equals
{b1 where b1 is Element of the carrier of a1: Class(the InternalRel of a1,b1) c= a2};
end;
:: ROUGHS_1:def 4
theorem
for b1 being non empty RelStr
for b2 being Element of bool the carrier of b1 holds
LAp b2 = {b3 where b3 is Element of the carrier of b1: Class(the InternalRel of b1,b3) c= b2};
:: ROUGHS_1:funcnot 4 => ROUGHS_1:func 4
definition
let a1 be non empty RelStr;
let a2 be Element of bool the carrier of a1;
func UAp A2 -> Element of bool the carrier of a1 equals
{b1 where b1 is Element of the carrier of a1: Class(the InternalRel of a1,b1) meets a2};
end;
:: ROUGHS_1:def 5
theorem
for b1 being non empty RelStr
for b2 being Element of bool the carrier of b1 holds
UAp b2 = {b3 where b3 is Element of the carrier of b1: Class(the InternalRel of b1,b3) meets b2};
:: ROUGHS_1:funcnot 5 => ROUGHS_1:func 5
definition
let a1 be non empty RelStr;
let a2 be Element of bool the carrier of a1;
func BndAp A2 -> Element of bool the carrier of a1 equals
(UAp a2) \ LAp a2;
end;
:: ROUGHS_1:def 6
theorem
for b1 being non empty RelStr
for b2 being Element of bool the carrier of b1 holds
BndAp b2 = (UAp b2) \ LAp b2;
:: ROUGHS_1:attrnot 4 => ROUGHS_1:attr 4
definition
let a1 be non empty RelStr;
let a2 be Element of bool the carrier of a1;
attr a2 is rough means
BndAp a2 <> {};
end;
:: ROUGHS_1:dfs 7
definiens
let a1 be non empty RelStr;
let a2 be Element of bool the carrier of a1;
To prove
a2 is rough
it is sufficient to prove
thus BndAp a2 <> {};
:: ROUGHS_1:def 7
theorem
for b1 being non empty RelStr
for b2 being Element of bool the carrier of b1 holds
b2 is rough(b1)
iff
BndAp b2 <> {};
:: ROUGHS_1:attrnot 5 => ROUGHS_1:attr 4
notation
let a1 be non empty RelStr;
let a2 be Element of bool the carrier of a1;
antonym exact for rough;
end;
:: ROUGHS_1:th 8
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being set
st b3 in LAp b2
holds Class(the InternalRel of b1,b3) c= b2;
:: ROUGHS_1:th 9
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1
st Class(the InternalRel of b1,b3) c= b2
holds b3 in LAp b2;
:: ROUGHS_1:th 10
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being set
st b3 in UAp b2
holds Class(the InternalRel of b1,b3) meets b2;
:: ROUGHS_1:th 11
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1
st Class(the InternalRel of b1,b3) meets b2
holds b3 in UAp b2;
:: ROUGHS_1:th 12
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
LAp b2 c= b2;
:: ROUGHS_1:th 13
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
b2 c= UAp b2;
:: ROUGHS_1:th 14
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
LAp b2 c= UAp b2;
:: ROUGHS_1:th 15
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
b2 is exact(b1)
iff
LAp b2 = b2;
:: ROUGHS_1:th 16
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
b2 is exact(b1)
iff
UAp b2 = b2;
:: ROUGHS_1:th 17
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
b2 = LAp b2
iff
b2 = UAp b2;
:: ROUGHS_1:th 18
theorem
for b1 being non empty with_tolerance RelStr holds
LAp {} b1 = {};
:: ROUGHS_1:th 19
theorem
for b1 being non empty with_tolerance RelStr holds
UAp {} b1 = {};
:: ROUGHS_1:th 20
theorem
for b1 being non empty with_tolerance RelStr holds
LAp [#] b1 = [#] b1;
:: ROUGHS_1:th 21
theorem
for b1 being non empty with_tolerance RelStr holds
UAp [#] b1 = [#] b1;
:: ROUGHS_1:th 22
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
LAp (b2 /\ b3) = (LAp b2) /\ LAp b3;
:: ROUGHS_1:th 23
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
UAp (b2 \/ b3) = (UAp b2) \/ UAp b3;
:: ROUGHS_1:th 24
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1
st b2 c= b3
holds LAp b2 c= LAp b3;
:: ROUGHS_1:th 25
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1
st b2 c= b3
holds UAp b2 c= UAp b3;
:: ROUGHS_1:th 26
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
(LAp b2) \/ LAp b3 c= LAp (b2 \/ b3);
:: ROUGHS_1:th 27
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
UAp (b2 /\ b3) c= (UAp b2) /\ UAp b3;
:: ROUGHS_1:th 28
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
LAp (b2 `) = (UAp b2) `;
:: ROUGHS_1:th 29
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
UAp (b2 `) = (LAp b2) `;
:: ROUGHS_1:th 30
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
UAp LAp UAp b2 = UAp b2;
:: ROUGHS_1:th 31
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
LAp UAp LAp b2 = LAp b2;
:: ROUGHS_1:th 32
theorem
for b1 being non empty with_tolerance RelStr
for b2 being Element of bool the carrier of b1 holds
BndAp b2 = BndAp (b2 `);
:: ROUGHS_1:th 33
theorem
for b1 being non empty with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
LAp LAp b2 = LAp b2;
:: ROUGHS_1:th 34
theorem
for b1 being non empty with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
LAp LAp b2 = UAp LAp b2;
:: ROUGHS_1:th 35
theorem
for b1 being non empty with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
UAp UAp b2 = UAp b2;
:: ROUGHS_1:th 36
theorem
for b1 being non empty with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
UAp UAp b2 = LAp UAp b2;
:: ROUGHS_1:exreg 6
registration
let a1 be non empty with_tolerance RelStr;
cluster exact Element of bool the carrier of a1;
end;
:: ROUGHS_1:funcreg 7
registration
let a1 be non empty with_equivalence RelStr;
let a2 be Element of bool the carrier of a1;
cluster LAp a2 -> exact;
end;
:: ROUGHS_1:funcreg 8
registration
let a1 be non empty with_equivalence RelStr;
let a2 be Element of bool the carrier of a1;
cluster UAp a2 -> exact;
end;
:: ROUGHS_1:th 37
theorem
for b1 being non empty with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3, b4 being set
st b3 in UAp b2 & [b3,b4] in the InternalRel of b1
holds b4 in UAp b2;
:: ROUGHS_1:exreg 7
registration
let a1 be non empty non diagonal with_equivalence RelStr;
cluster rough Element of bool the carrier of a1;
end;
:: ROUGHS_1:modenot 3 => ROUGHS_1:mode 1
definition
let a1 be non empty with_equivalence RelStr;
let a2 be Element of bool the carrier of a1;
mode RoughSet of A2 means
it = [LAp a2,UAp a2];
end;
:: ROUGHS_1:dfs 8
definiens
let a1 be non empty with_equivalence RelStr;
let a2 be Element of bool the carrier of a1;
let a3 be set;
To prove
a3 is RoughSet of a2
it is sufficient to prove
thus a3 = [LAp a2,UAp a2];
:: ROUGHS_1:def 8
theorem
for b1 being non empty with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3 being set holds
b3 is RoughSet of b2
iff
b3 = [LAp b2,UAp b2];
:: ROUGHS_1:funcreg 9
registration
let a1 be non empty finite with_tolerance RelStr;
let a2 be Element of the carrier of a1;
cluster Card Class(the InternalRel of a1,a2) -> non empty cardinal;
end;
:: ROUGHS_1:funcnot 6 => ROUGHS_1:func 6
definition
let a1 be non empty finite with_tolerance RelStr;
let a2 be Element of bool the carrier of a1;
func MemberFunc(A2,A1) -> Function-like quasi_total Relation of the carrier of a1,REAL means
for b1 being Element of the carrier of a1 holds
it . b1 = (card (a2 /\ Class(the InternalRel of a1,b1))) / card Class(the InternalRel of a1,b1);
end;
:: ROUGHS_1:def 9
theorem
for b1 being non empty finite with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being Function-like quasi_total Relation of the carrier of b1,REAL holds
b3 = MemberFunc(b2,b1)
iff
for b4 being Element of the carrier of b1 holds
b3 . b4 = (card (b2 /\ Class(the InternalRel of b1,b4))) / card Class(the InternalRel of b1,b4);
:: ROUGHS_1:th 38
theorem
for b1 being non empty finite with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1 holds
0 <= (MemberFunc(b2,b1)) . b3 & (MemberFunc(b2,b1)) . b3 <= 1;
:: ROUGHS_1:th 39
theorem
for b1 being non empty finite with_tolerance RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1 holds
(MemberFunc(b2,b1)) . b3 in [.0,1.];
:: ROUGHS_1:th 40
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1 holds
(MemberFunc(b2,b1)) . b3 = 1
iff
b3 in LAp b2;
:: ROUGHS_1:th 41
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1 holds
(MemberFunc(b2,b1)) . b3 = 0
iff
b3 in (UAp b2) `;
:: ROUGHS_1:th 42
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1 holds
0 < (MemberFunc(b2,b1)) . b3 & (MemberFunc(b2,b1)) . b3 < 1
iff
b3 in BndAp b2;
:: ROUGHS_1:th 43
theorem
for b1 being non empty discrete with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
b2 is exact(b1);
:: ROUGHS_1:condreg 7
registration
let a1 be non empty discrete with_equivalence RelStr;
cluster -> exact (Element of bool the carrier of a1);
end;
:: ROUGHS_1:th 44
theorem
for b1 being non empty finite discrete with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
MemberFunc(b2,b1) = chi(b2,the carrier of b1);
:: ROUGHS_1:th 45
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3, b4 being set
st [b3,b4] in the InternalRel of b1
holds (MemberFunc(b2,b1)) . b3 = (MemberFunc(b2,b1)) . b4;
:: ROUGHS_1:th 46
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1
for b3 being Element of the carrier of b1 holds
(MemberFunc(b2 `,b1)) . b3 = 1 - ((MemberFunc(b2,b1)) . b3);
:: ROUGHS_1:th 47
theorem
for b1 being non empty finite with_equivalence RelStr
for b2, b3 being Element of bool the carrier of b1
for b4 being Element of the carrier of b1
st b2 c= b3
holds (MemberFunc(b2,b1)) . b4 <= (MemberFunc(b3,b1)) . b4;
:: ROUGHS_1:th 48
theorem
for b1 being non empty finite with_equivalence RelStr
for b2, b3 being Element of bool the carrier of b1
for b4 being Element of the carrier of b1 holds
(MemberFunc(b2,b1)) . b4 <= (MemberFunc(b2 \/ b3,b1)) . b4;
:: ROUGHS_1:th 49
theorem
for b1 being non empty finite with_equivalence RelStr
for b2, b3 being Element of bool the carrier of b1
for b4 being Element of the carrier of b1 holds
(MemberFunc(b2 /\ b3,b1)) . b4 <= (MemberFunc(b2,b1)) . b4;
:: ROUGHS_1:th 50
theorem
for b1 being non empty finite with_equivalence RelStr
for b2, b3 being Element of bool the carrier of b1
for b4 being Element of the carrier of b1 holds
max((MemberFunc(b2,b1)) . b4,(MemberFunc(b3,b1)) . b4) <= (MemberFunc(b2 \/ b3,b1)) . b4;
:: ROUGHS_1:th 51
theorem
for b1 being non empty finite with_equivalence RelStr
for b2, b3 being Element of bool the carrier of b1
for b4 being Element of the carrier of b1
st b2 misses b3
holds (MemberFunc(b2 \/ b3,b1)) . b4 = ((MemberFunc(b2,b1)) . b4) + ((MemberFunc(b3,b1)) . b4);
:: ROUGHS_1:th 52
theorem
for b1 being non empty finite with_equivalence RelStr
for b2, b3 being Element of bool the carrier of b1
for b4 being Element of the carrier of b1 holds
(MemberFunc(b2 /\ b3,b1)) . b4 <= min((MemberFunc(b2,b1)) . b4,(MemberFunc(b3,b1)) . b4);
:: ROUGHS_1:funcnot 7 => ROUGHS_1:func 7
definition
let a1 be non empty finite with_tolerance RelStr;
let a2 be FinSequence of bool the carrier of a1;
let a3 be Element of the carrier of a1;
func FinSeqM(A3,A2) -> FinSequence of REAL means
dom it = dom a2 &
(for b1 being natural set
st b1 in dom a2
holds it . b1 = (MemberFunc(a2 . b1,a1)) . a3);
end;
:: ROUGHS_1:def 10
theorem
for b1 being non empty finite with_tolerance RelStr
for b2 being FinSequence of bool the carrier of b1
for b3 being Element of the carrier of b1
for b4 being FinSequence of REAL holds
b4 = FinSeqM(b3,b2)
iff
dom b4 = dom b2 &
(for b5 being natural set
st b5 in dom b2
holds b4 . b5 = (MemberFunc(b2 . b5,b1)) . b3);
:: ROUGHS_1:th 53
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being FinSequence of bool the carrier of b1
for b3 being Element of the carrier of b1
for b4 being Element of bool the carrier of b1 holds
FinSeqM(b3,b2 ^ <*b4*>) = (FinSeqM(b3,b2)) ^ <*(MemberFunc(b4,b1)) . b3*>;
:: ROUGHS_1:th 54
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of the carrier of b1 holds
(MemberFunc({} b1,b1)) . b2 = 0;
:: ROUGHS_1:th 55
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of the carrier of b1
for b3 being disjoint_valued FinSequence of bool the carrier of b1 holds
(MemberFunc(Union b3,b1)) . b2 = Sum FinSeqM(b2,b3);
:: ROUGHS_1:th 56
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
LAp b2 = {b3 where b3 is Element of the carrier of b1: (MemberFunc(b2,b1)) . b3 = 1};
:: ROUGHS_1:th 57
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
UAp b2 = {b3 where b3 is Element of the carrier of b1: 0 < (MemberFunc(b2,b1)) . b3};
:: ROUGHS_1:th 58
theorem
for b1 being non empty finite with_equivalence RelStr
for b2 being Element of bool the carrier of b1 holds
BndAp b2 = {b3 where b3 is Element of the carrier of b1: 0 < (MemberFunc(b2,b1)) . b3 & (MemberFunc(b2,b1)) . b3 < 1};
:: ROUGHS_1:prednot 1 => ROUGHS_1:pred 1
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 _c= A3 means
LAp a2 c= LAp a3;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 _c= a2;
end;
:: ROUGHS_1:dfs 11
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 _c= a3
it is sufficient to prove
thus LAp a2 c= LAp a3;
:: ROUGHS_1:def 11
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 _c= b3
iff
LAp b2 c= LAp b3;
:: ROUGHS_1:prednot 2 => ROUGHS_1:pred 2
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 c=^ A3 means
UAp a2 c= UAp a3;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 c=^ a2;
end;
:: ROUGHS_1:dfs 12
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 c=^ a3
it is sufficient to prove
thus UAp a2 c= UAp a3;
:: ROUGHS_1:def 12
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 c=^ b3
iff
UAp b2 c= UAp b3;
:: ROUGHS_1:prednot 3 => ROUGHS_1:pred 3
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 _c=^ A3 means
a2 _c= a3 & a2 c=^ a3;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 _c=^ a2;
end;
:: ROUGHS_1:dfs 13
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 _c=^ a3
it is sufficient to prove
thus a2 _c= a3 & a2 c=^ a3;
:: ROUGHS_1:def 13
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 _c=^ b3
iff
b2 _c= b3 & b2 c=^ b3;
:: ROUGHS_1:th 59
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3, b4 being Element of bool the carrier of b1
st b2 _c= b3 & b3 _c= b4
holds b2 _c= b4;
:: ROUGHS_1:th 60
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3, b4 being Element of bool the carrier of b1
st b2 c=^ b3 & b3 c=^ b4
holds b2 c=^ b4;
:: ROUGHS_1:th 61
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3, b4 being Element of bool the carrier of b1
st b2 _c=^ b3 & b3 _c=^ b4
holds b2 _c=^ b4;
:: ROUGHS_1:prednot 4 => ROUGHS_1:pred 4
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 _= A3 means
LAp a2 = LAp a3;
symmetry;
:: for a1 being non empty with_tolerance RelStr
:: for a2, a3 being Element of bool the carrier of a1
:: st a2 _= a3
:: holds a3 _= a2;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 _= a2;
end;
:: ROUGHS_1:dfs 14
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 _= a3
it is sufficient to prove
thus LAp a2 = LAp a3;
:: ROUGHS_1:def 14
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 _= b3
iff
LAp b2 = LAp b3;
:: ROUGHS_1:prednot 5 => ROUGHS_1:pred 5
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 =^ A3 means
UAp a2 = UAp a3;
symmetry;
:: for a1 being non empty with_tolerance RelStr
:: for a2, a3 being Element of bool the carrier of a1
:: st a2 =^ a3
:: holds a3 =^ a2;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 =^ a2;
end;
:: ROUGHS_1:dfs 15
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 =^ a3
it is sufficient to prove
thus UAp a2 = UAp a3;
:: ROUGHS_1:def 15
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 =^ b3
iff
UAp b2 = UAp b3;
:: ROUGHS_1:prednot 6 => ROUGHS_1:pred 6
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 _=^ A3 means
LAp a2 = LAp a3 & UAp a2 = UAp a3;
symmetry;
:: for a1 being non empty with_tolerance RelStr
:: for a2, a3 being Element of bool the carrier of a1
:: st a2 _=^ a3
:: holds a3 _=^ a2;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 _=^ a2;
end;
:: ROUGHS_1:dfs 16
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 _=^ a3
it is sufficient to prove
thus LAp a2 = LAp a3 & UAp a2 = UAp a3;
:: ROUGHS_1:def 16
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 _=^ b3
iff
LAp b2 = LAp b3 & UAp b2 = UAp b3;
:: ROUGHS_1:prednot 7 => ROUGHS_1:pred 4
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 _= A3 means
a2 _c= a3 & a3 _c= a2;
symmetry;
:: for a1 being non empty with_tolerance RelStr
:: for a2, a3 being Element of bool the carrier of a1
:: st a2 _= a3
:: holds a3 _= a2;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 _= a2;
end;
:: ROUGHS_1:dfs 17
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 _= a3
it is sufficient to prove
thus a2 _c= a3 & a3 _c= a2;
:: ROUGHS_1:def 17
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 _= b3
iff
b2 _c= b3 & b3 _c= b2;
:: ROUGHS_1:prednot 8 => ROUGHS_1:pred 5
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 =^ A3 means
a2 c=^ a3 & a3 c=^ a2;
symmetry;
:: for a1 being non empty with_tolerance RelStr
:: for a2, a3 being Element of bool the carrier of a1
:: st a2 =^ a3
:: holds a3 =^ a2;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 =^ a2;
end;
:: ROUGHS_1:dfs 18
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 =^ a3
it is sufficient to prove
thus a2 c=^ a3 & a3 c=^ a2;
:: ROUGHS_1:def 18
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 =^ b3
iff
b2 c=^ b3 & b3 c=^ b2;
:: ROUGHS_1:prednot 9 => ROUGHS_1:pred 6
definition
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
pred A2 _=^ A3 means
a2 _= a3 & a2 =^ a3;
symmetry;
:: for a1 being non empty with_tolerance RelStr
:: for a2, a3 being Element of bool the carrier of a1
:: st a2 _=^ a3
:: holds a3 _=^ a2;
reflexivity;
:: for a1 being non empty with_tolerance RelStr
:: for a2 being Element of bool the carrier of a1 holds
:: a2 _=^ a2;
end;
:: ROUGHS_1:dfs 19
definiens
let a1 be non empty with_tolerance RelStr;
let a2, a3 be Element of bool the carrier of a1;
To prove
a2 _=^ a3
it is sufficient to prove
thus a2 _= a3 & a2 =^ a3;
:: ROUGHS_1:def 19
theorem
for b1 being non empty with_tolerance RelStr
for b2, b3 being Element of bool the carrier of b1 holds
b2 _=^ b3
iff
b2 _= b3 & b2 =^ b3;