Step
*
1
of Lemma
b_all-squash-exists-bag
1. A : Type
2. B : Type
3. P : A ⟶ B ⟶ ℙ
4. b_all(A;[];x.↓∃y:B. P[x;y])
⊢ ↓∃bs:bag(A × B). ((bag-map(λx.(fst(x));bs) = [] ∈ bag(A)) ∧ b_all(A × B;bs;x.let a,b = x in ↓P[a;b]))
BY
{ (D 0 THEN InstConcl [⌜[]⌝]⋅ THEN Auto THEN D 0 THEN Auto THEN BagMemberD (-1) THEN Auto) }
Latex:
Latex:
1. A : Type
2. B : Type
3. P : A {}\mrightarrow{} B {}\mrightarrow{} \mBbbP{}
4. b\_all(A;[];x.\mdownarrow{}\mexists{}y:B. P[x;y])
\mvdash{} \mdownarrow{}\mexists{}bs:bag(A \mtimes{} B). ((bag-map(\mlambda{}x.(fst(x));bs) = []) \mwedge{} b\_all(A \mtimes{} B;bs;x.let a,b = x in \mdownarrow{}P[a;b]))
By
Latex:
(D 0 THEN InstConcl [\mkleeneopen{}[]\mkleeneclose{}]\mcdot{} THEN Auto THEN D 0 THEN Auto THEN BagMemberD (-1) THEN Auto)
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