Step
*
2
2
of Lemma
bag-drop-append
1. T : Type
2. eq : EqDecider(T)
3. x : T
4. ∀bs:bag(T). ((bs = ({x} + bag-drop(eq;bs;x)) ∈ bag(T)) ∨ ((¬x ↓∈ bs) ∧ (bs = bag-drop(eq;bs;x) ∈ bag(T))))
5. bs : bag(T)
6. cs : bag(T)
7. ¬x ↓∈ bs + cs
8. (bs + cs) = bag-drop(eq;bs + cs;x) ∈ bag(T)
⊢ bag-drop(eq;bs + cs;x) = (bs + bag-drop(eq;cs;x)) ∈ bag(T)
BY
{ ((D 4 With ⌜cs⌝ THENA Auto) THEN D -1 THEN Auto) }
1
1. T : Type
2. eq : EqDecider(T)
3. x : T
4. bs : bag(T)
5. cs : bag(T)
6. ¬x ↓∈ bs + cs
7. (bs + cs) = bag-drop(eq;bs + cs;x) ∈ bag(T)
8. cs = ({x} + bag-drop(eq;cs;x)) ∈ bag(T)
⊢ bag-drop(eq;bs + cs;x) = (bs + bag-drop(eq;cs;x)) ∈ bag(T)
Latex:
Latex:
1. T : Type
2. eq : EqDecider(T)
3. x : T
4. \mforall{}bs:bag(T). ((bs = (\{x\} + bag-drop(eq;bs;x))) \mvee{} ((\mneg{}x \mdownarrow{}\mmember{} bs) \mwedge{} (bs = bag-drop(eq;bs;x))))
5. bs : bag(T)
6. cs : bag(T)
7. \mneg{}x \mdownarrow{}\mmember{} bs + cs
8. (bs + cs) = bag-drop(eq;bs + cs;x)
\mvdash{} bag-drop(eq;bs + cs;x) = (bs + bag-drop(eq;cs;x))
By
Latex:
((D 4 With \mkleeneopen{}cs\mkleeneclose{} THENA Auto) THEN D -1 THEN Auto)
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