Step
*
2
1
1
of Lemma
bag-member-filter-implies2
1. T : Type
2. x : T
3. bs : bag(T)
4. ∀b1:T List. ∀x1:{x:T| x ↓∈ b1} ⟶ 𝔹. ∀x:T. (x ↓∈ [x∈b1|x1[x]]
⇒ ((Ax ∈ x ↓∈ b1) ∧ (Ax ∈ ↑x1[x])))
⊢ λp.<Ax, Ax> ∈ ∀P:{x:T| x ↓∈ bs} ⟶ 𝔹. x ↓∈ bs ∧ (↑P[x]) supposing x ↓∈ [x∈bs|P[x]]
BY
{ TACTIC:BagD 3 }
1
.....wf.....
1. T : Type
2. x : T
3. bs : as,bs:T List//permutation(T;as;bs)
4. ∀b1:T List. ∀x1:{x:T| x ↓∈ b1} ⟶ 𝔹. ∀x:T. (x ↓∈ [x∈b1|x1[x]]
⇒ ((Ax ∈ x ↓∈ b1) ∧ (Ax ∈ ↑x1[x])))
⊢ ∀P:{x:T| x ↓∈ bs} ⟶ 𝔹. x ↓∈ bs ∧ (↑P[x]) supposing x ↓∈ [x∈bs|P[x]] ∈ Type
2
1. T : Type
2. x : T
3. ∀b1:T List. ∀x1:{x:T| x ↓∈ b1} ⟶ 𝔹. ∀x:T. (x ↓∈ [x∈b1|x1[x]]
⇒ ((Ax ∈ x ↓∈ b1) ∧ (Ax ∈ ↑x1[x])))
4. as : T List
5. bs : T List
6. permutation(T;as;bs)
⊢ (λp.<Ax, Ax>) = (λp.<Ax, Ax>) ∈ (∀P:{x:T| x ↓∈ as} ⟶ 𝔹. x ↓∈ as ∧ (↑P[x]) supposing x ↓∈ [x∈as|P[x]])
Latex:
Latex:
1. T : Type
2. x : T
3. bs : bag(T)
4. \mforall{}b1:T List. \mforall{}x1:\{x:T| x \mdownarrow{}\mmember{} b1\} {}\mrightarrow{} \mBbbB{}. \mforall{}x:T.
(x \mdownarrow{}\mmember{} [x\mmember{}b1|x1[x]] {}\mRightarrow{} ((Ax \mmember{} x \mdownarrow{}\mmember{} b1) \mwedge{} (Ax \mmember{} \muparrow{}x1[x])))
\mvdash{} \mlambda{}p.<Ax, Ax> \mmember{} \mforall{}P:\{x:T| x \mdownarrow{}\mmember{} bs\} {}\mrightarrow{} \mBbbB{}. x \mdownarrow{}\mmember{} bs \mwedge{} (\muparrow{}P[x]) supposing x \mdownarrow{}\mmember{} [x\mmember{}bs|P[x]]
By
Latex:
TACTIC:BagD 3
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