Step
*
of Lemma
bag-append-equal-bag-rep
No Annotations
∀[T:Type]. ∀[x:T]. ∀[n:ℕ]. ∀[a,b:bag(T)].
uiff((a + b) = bag-rep(n;x) ∈ bag(T);(n = (#(a) + #(b)) ∈ ℤ)
∧ (a = bag-rep(#(a);x) ∈ bag(T))
∧ (b = bag-rep(#(b);x) ∈ bag(T)))
BY
{ TACTIC:Auto }
1
1. T : Type
2. x : T
3. n : ℕ
4. a : bag(T)
5. b : bag(T)
6. (a + b) = bag-rep(n;x) ∈ bag(T)
⊢ n = (#(a) + #(b)) ∈ ℤ
2
1. T : Type
2. x : T
3. n : ℕ
4. a : bag(T)
5. b : bag(T)
6. (a + b) = bag-rep(n;x) ∈ bag(T)
⊢ a = bag-rep(#(a);x) ∈ bag(T)
3
1. T : Type
2. x : T
3. n : ℕ
4. a : bag(T)
5. b : bag(T)
6. (a + b) = bag-rep(n;x) ∈ bag(T)
⊢ b = bag-rep(#(b);x) ∈ bag(T)
4
1. T : Type
2. x : T
3. n : ℕ
4. a : bag(T)
5. b : bag(T)
6. n = (#(a) + #(b)) ∈ ℤ
7. a = bag-rep(#(a);x) ∈ bag(T)
8. b = bag-rep(#(b);x) ∈ bag(T)
⊢ (a + b) = bag-rep(n;x) ∈ bag(T)
Latex:
Latex:
No Annotations
\mforall{}[T:Type]. \mforall{}[x:T]. \mforall{}[n:\mBbbN{}]. \mforall{}[a,b:bag(T)].
uiff((a + b) = bag-rep(n;x);(n = (\#(a) + \#(b))) \mwedge{} (a = bag-rep(\#(a);x)) \mwedge{} (b = bag-rep(\#(b);x)))
By
Latex:
TACTIC:Auto
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