lsppropd

Description: If two structures have the same components (properties), they have the same span function. (Contributed by Mario Carneiro, 9-Feb-2015) (Revised by Mario Carneiro, 14-Jun-2015) (Revised by AV, 24-Apr-2024)

	Ref	Expression
Hypotheses	lsspropd.b1	⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐾 ) )
	lsspropd.b2	⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐿 ) )
	lsspropd.w	⊢ ( 𝜑 → 𝐵 ⊆ 𝑊 )
	lsspropd.p	⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑊 ∧ 𝑦 ∈ 𝑊 ) ) → ( 𝑥 ( +_g ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( +_g ‘ 𝐿 ) 𝑦 ) )
	lsspropd.s1	⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( ·_𝑠 ‘ 𝐾 ) 𝑦 ) ∈ 𝑊 )
	lsspropd.s2	⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( ·_𝑠 ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( ·_𝑠 ‘ 𝐿 ) 𝑦 ) )
	lsspropd.p1	⊢ ( 𝜑 → 𝑃 = ( Base ‘ ( Scalar ‘ 𝐾 ) ) )
	lsspropd.p2	⊢ ( 𝜑 → 𝑃 = ( Base ‘ ( Scalar ‘ 𝐿 ) ) )
	lsppropd.v1	⊢ ( 𝜑 → 𝐾 ∈ 𝑋 )
	lsppropd.v2	⊢ ( 𝜑 → 𝐿 ∈ 𝑌 )
Assertion	lsppropd	⊢ ( 𝜑 → ( LSpan ‘ 𝐾 ) = ( LSpan ‘ 𝐿 ) )

Proof

Step	Hyp	Ref	Expression
1		lsspropd.b1	⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐾 ) )
2		lsspropd.b2	⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐿 ) )
3		lsspropd.w	⊢ ( 𝜑 → 𝐵 ⊆ 𝑊 )
4		lsspropd.p	⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑊 ∧ 𝑦 ∈ 𝑊 ) ) → ( 𝑥 ( +_g ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( +_g ‘ 𝐿 ) 𝑦 ) )
5		lsspropd.s1	⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( ·_𝑠 ‘ 𝐾 ) 𝑦 ) ∈ 𝑊 )
6		lsspropd.s2	⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( ·_𝑠 ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( ·_𝑠 ‘ 𝐿 ) 𝑦 ) )
7		lsspropd.p1	⊢ ( 𝜑 → 𝑃 = ( Base ‘ ( Scalar ‘ 𝐾 ) ) )
8		lsspropd.p2	⊢ ( 𝜑 → 𝑃 = ( Base ‘ ( Scalar ‘ 𝐿 ) ) )
9		lsppropd.v1	⊢ ( 𝜑 → 𝐾 ∈ 𝑋 )
10		lsppropd.v2	⊢ ( 𝜑 → 𝐿 ∈ 𝑌 )
11	1 2	eqtr3d	⊢ ( 𝜑 → ( Base ‘ 𝐾 ) = ( Base ‘ 𝐿 ) )
12	11	pweqd	⊢ ( 𝜑 → 𝒫 ( Base ‘ 𝐾 ) = 𝒫 ( Base ‘ 𝐿 ) )
13	1 2 3 4 5 6 7 8	lsspropd	⊢ ( 𝜑 → ( LSubSp ‘ 𝐾 ) = ( LSubSp ‘ 𝐿 ) )
14	13	rabeqdv	⊢ ( 𝜑 → { 𝑡 ∈ ( LSubSp ‘ 𝐾 ) ∣ 𝑠 ⊆ 𝑡 } = { 𝑡 ∈ ( LSubSp ‘ 𝐿 ) ∣ 𝑠 ⊆ 𝑡 } )
15	14	inteqd	⊢ ( 𝜑 → ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐾 ) ∣ 𝑠 ⊆ 𝑡 } = ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐿 ) ∣ 𝑠 ⊆ 𝑡 } )
16	12 15	mpteq12dv	⊢ ( 𝜑 → ( 𝑠 ∈ 𝒫 ( Base ‘ 𝐾 ) ↦ ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐾 ) ∣ 𝑠 ⊆ 𝑡 } ) = ( 𝑠 ∈ 𝒫 ( Base ‘ 𝐿 ) ↦ ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐿 ) ∣ 𝑠 ⊆ 𝑡 } ) )
17		eqid	⊢ ( Base ‘ 𝐾 ) = ( Base ‘ 𝐾 )
18		eqid	⊢ ( LSubSp ‘ 𝐾 ) = ( LSubSp ‘ 𝐾 )
19		eqid	⊢ ( LSpan ‘ 𝐾 ) = ( LSpan ‘ 𝐾 )
20	17 18 19	lspfval	⊢ ( 𝐾 ∈ 𝑋 → ( LSpan ‘ 𝐾 ) = ( 𝑠 ∈ 𝒫 ( Base ‘ 𝐾 ) ↦ ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐾 ) ∣ 𝑠 ⊆ 𝑡 } ) )
21	9 20	syl	⊢ ( 𝜑 → ( LSpan ‘ 𝐾 ) = ( 𝑠 ∈ 𝒫 ( Base ‘ 𝐾 ) ↦ ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐾 ) ∣ 𝑠 ⊆ 𝑡 } ) )
22		eqid	⊢ ( Base ‘ 𝐿 ) = ( Base ‘ 𝐿 )
23		eqid	⊢ ( LSubSp ‘ 𝐿 ) = ( LSubSp ‘ 𝐿 )
24		eqid	⊢ ( LSpan ‘ 𝐿 ) = ( LSpan ‘ 𝐿 )
25	22 23 24	lspfval	⊢ ( 𝐿 ∈ 𝑌 → ( LSpan ‘ 𝐿 ) = ( 𝑠 ∈ 𝒫 ( Base ‘ 𝐿 ) ↦ ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐿 ) ∣ 𝑠 ⊆ 𝑡 } ) )
26	10 25	syl	⊢ ( 𝜑 → ( LSpan ‘ 𝐿 ) = ( 𝑠 ∈ 𝒫 ( Base ‘ 𝐿 ) ↦ ∩ { 𝑡 ∈ ( LSubSp ‘ 𝐿 ) ∣ 𝑠 ⊆ 𝑡 } ) )
27	16 21 26	3eqtr4d	⊢ ( 𝜑 → ( LSpan ‘ 𝐾 ) = ( LSpan ‘ 𝐿 ) )

Metamath Proof Explorer

Theorem lsppropd

Proof