Theorem smucl

Description: The product of two sequences is a sequence. (Contributed by Mario Carneiro, 19-Sep-2016)

	Ref	Expression
Assertion	smucl	⊢ ( ( 𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0 ) → ( 𝐴 smul 𝐵 ) ⊆ ℕ0 )

Proof

Step	Hyp	Ref	Expression
1		simpl	⊢ ( ( 𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0 ) → 𝐴 ⊆ ℕ0 )
2		simpr	⊢ ( ( 𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0 ) → 𝐵 ⊆ ℕ0 )
3		eqid	⊢ seq 0 ( ( 𝑝 ∈ 𝒫 ℕ0 , 𝑚 ∈ ℕ0 ↦ ( 𝑝 sadd { 𝑛 ∈ ℕ0 ∣ ( 𝑚 ∈ 𝐴 ∧ ( 𝑛 − 𝑚 ) ∈ 𝐵 ) } ) ) , ( 𝑛 ∈ ℕ0 ↦ if ( 𝑛 = 0 , ∅ , ( 𝑛 − 1 ) ) ) ) = seq 0 ( ( 𝑝 ∈ 𝒫 ℕ0 , 𝑚 ∈ ℕ0 ↦ ( 𝑝 sadd { 𝑛 ∈ ℕ0 ∣ ( 𝑚 ∈ 𝐴 ∧ ( 𝑛 − 𝑚 ) ∈ 𝐵 ) } ) ) , ( 𝑛 ∈ ℕ0 ↦ if ( 𝑛 = 0 , ∅ , ( 𝑛 − 1 ) ) ) )
4	1 2 3	smufval	⊢ ( ( 𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0 ) → ( 𝐴 smul 𝐵 ) = { 𝑘 ∈ ℕ0 ∣ 𝑘 ∈ ( seq 0 ( ( 𝑝 ∈ 𝒫 ℕ0 , 𝑚 ∈ ℕ0 ↦ ( 𝑝 sadd { 𝑛 ∈ ℕ0 ∣ ( 𝑚 ∈ 𝐴 ∧ ( 𝑛 − 𝑚 ) ∈ 𝐵 ) } ) ) , ( 𝑛 ∈ ℕ0 ↦ if ( 𝑛 = 0 , ∅ , ( 𝑛 − 1 ) ) ) ) ‘ ( 𝑘 + 1 ) ) } )
5		ssrab2	⊢ { 𝑘 ∈ ℕ0 ∣ 𝑘 ∈ ( seq 0 ( ( 𝑝 ∈ 𝒫 ℕ0 , 𝑚 ∈ ℕ0 ↦ ( 𝑝 sadd { 𝑛 ∈ ℕ0 ∣ ( 𝑚 ∈ 𝐴 ∧ ( 𝑛 − 𝑚 ) ∈ 𝐵 ) } ) ) , ( 𝑛 ∈ ℕ0 ↦ if ( 𝑛 = 0 , ∅ , ( 𝑛 − 1 ) ) ) ) ‘ ( 𝑘 + 1 ) ) } ⊆ ℕ0
6	4 5	eqsstrdi	⊢ ( ( 𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0 ) → ( 𝐴 smul 𝐵 ) ⊆ ℕ0 )