Step
*
1
of Lemma
fset-subtype
.....antecedent.....
1. A : Type
2. B : Type
3. A ⊆r B
4. x : Base
5. x1 : Base
6. x = x1 ∈ pertype(λx,y. ((x ∈ A List) ∧ (y ∈ A List) ∧ set-equal(A;x;y)))
7. x ∈ A List
8. x1 ∈ A List
9. set-equal(A;x;x1)
⊢ set-equal(B;x;x1)
BY
{ (ParallelLast THEN Auto THEN D -1 THEN ⌜ExRepD⌝⋅) }
1
1. A : Type
2. B : Type
3. A ⊆r B
4. x : Base
5. x1 : Base
6. x = x1 ∈ pertype(λx,y. ((x ∈ A List) ∧ (y ∈ A List) ∧ set-equal(A;x;y)))
7. x ∈ A List
8. x1 ∈ A List
9. ∀t:A. ((t ∈ x)
⇐⇒ (t ∈ x1))
10. t : B
11. i : ℕ
12. i < ||x||
13. t = x[i] ∈ B
⊢ (t ∈ x1)
2
1. A : Type
2. B : Type
3. A ⊆r B
4. x : Base
5. x1 : Base
6. x = x1 ∈ pertype(λx,y. ((x ∈ A List) ∧ (y ∈ A List) ∧ set-equal(A;x;y)))
7. x ∈ A List
8. x1 ∈ A List
9. ∀t:A. ((t ∈ x)
⇐⇒ (t ∈ x1))
10. t : B
11. i : ℕ
12. i < ||x1||
13. t = x1[i] ∈ B
⊢ (t ∈ x)
Latex:
Latex:
.....antecedent.....
1. A : Type
2. B : Type
3. A \msubseteq{}r B
4. x : Base
5. x1 : Base
6. x = x1
7. x \mmember{} A List
8. x1 \mmember{} A List
9. set-equal(A;x;x1)
\mvdash{} set-equal(B;x;x1)
By
Latex:
(ParallelLast THEN Auto THEN D -1 THEN \mkleeneopen{}ExRepD\mkleeneclose{}\mcdot{})
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