Step
*
of Lemma
bag-extensionality1
∀[T:Type]. ∀[eq:T ⟶ T ⟶ 𝔹].
∀[as,bs:bag(T)]. uiff(as = bs ∈ bag(T);∀x:T. (#([y∈as|eq x y]) = #([y∈bs|eq x y]) ∈ ℤ))
supposing ∀[x,y:T]. (↑(eq x y)
⇐⇒ x = y ∈ T)
BY
{ (Auto THEN Try (RepeatFor 2 ((EqCD THEN Auto))⋅)) }
1
1. T : Type
2. eq : T ⟶ T ⟶ 𝔹
3. ∀[x,y:T]. (↑(eq x y)
⇐⇒ x = y ∈ T)
4. as : bag(T)
5. bs : bag(T)
6. ∀x:T. (#([y∈as|eq x y]) = #([y∈bs|eq x y]) ∈ ℤ)
⊢ as = bs ∈ bag(T)
Latex:
Latex:
\mforall{}[T:Type]. \mforall{}[eq:T {}\mrightarrow{} T {}\mrightarrow{} \mBbbB{}].
\mforall{}[as,bs:bag(T)]. uiff(as = bs;\mforall{}x:T. (\#([y\mmember{}as|eq x y]) = \#([y\mmember{}bs|eq x y])))
supposing \mforall{}[x,y:T]. (\muparrow{}(eq x y) \mLeftarrow{}{}\mRightarrow{} x = y)
By
Latex:
(Auto THEN Try (RepeatFor 2 ((EqCD THEN Auto))\mcdot{}))
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