Step
*
1
1
1
of Lemma
bag-remove1-property
.....assertion.....
1. [T] : Type
2. eq : EqDecider(T)
3. x : T
4. bs : bag(T)
5. ∀x:T. ∀bs:bag(T). Dec(x ↓∈ bs)
6. x ↓∈ bs
⊢ (↑isl(bag-remove1(eq;bs;x))) ∧ (bs = ({x} + outl(bag-remove1(eq;bs;x))) ∈ bag(T))
BY
{ Assert ⌜↓(↑isl(bag-remove1(eq;bs;x))) ∧ (bs = ({x} + outl(bag-remove1(eq;bs;x))) ∈ bag(T))⌝⋅ }
1
.....assertion.....
1. [T] : Type
2. eq : EqDecider(T)
3. x : T
4. bs : bag(T)
5. ∀x:T. ∀bs:bag(T). Dec(x ↓∈ bs)
6. x ↓∈ bs
⊢ ↓(↑isl(bag-remove1(eq;bs;x))) ∧ (bs = ({x} + outl(bag-remove1(eq;bs;x))) ∈ bag(T))
2
1. [T] : Type
2. eq : EqDecider(T)
3. x : T
4. bs : bag(T)
5. ∀x:T. ∀bs:bag(T). Dec(x ↓∈ bs)
6. x ↓∈ bs
7. ↓(↑isl(bag-remove1(eq;bs;x))) ∧ (bs = ({x} + outl(bag-remove1(eq;bs;x))) ∈ bag(T))
⊢ (↑isl(bag-remove1(eq;bs;x))) ∧ (bs = ({x} + outl(bag-remove1(eq;bs;x))) ∈ bag(T))
Latex:
Latex:
.....assertion.....
1. [T] : Type
2. eq : EqDecider(T)
3. x : T
4. bs : bag(T)
5. \mforall{}x:T. \mforall{}bs:bag(T). Dec(x \mdownarrow{}\mmember{} bs)
6. x \mdownarrow{}\mmember{} bs
\mvdash{} (\muparrow{}isl(bag-remove1(eq;bs;x))) \mwedge{} (bs = (\{x\} + outl(bag-remove1(eq;bs;x))))
By
Latex:
Assert \mkleeneopen{}\mdownarrow{}(\muparrow{}isl(bag-remove1(eq;bs;x))) \mwedge{} (bs = (\{x\} + outl(bag-remove1(eq;bs;x))))\mkleeneclose{}\mcdot{}
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