Metamath Proof Explorer


Theorem 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 + K y = x + L y
lsspropd.s1 φ x P y B x K y W
lsspropd.s2 φ x P y B x K y = x 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 + K y = x + L y
5 lsspropd.s1 φ x P y B x K y W
6 lsspropd.s2 φ x P y B x K y = x 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