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 φB=BaseK
lsspropd.b2 φB=BaseL
lsspropd.w φBW
lsspropd.p φxWyWx+Ky=x+Ly
lsspropd.s1 φxPyBxKyW
lsspropd.s2 φxPyBxKy=xLy
lsspropd.p1 φP=BaseScalarK
lsspropd.p2 φP=BaseScalarL
lsppropd.v1 φKX
lsppropd.v2 φLY
Assertion lsppropd φLSpanK=LSpanL

Proof

Step Hyp Ref Expression
1 lsspropd.b1 φB=BaseK
2 lsspropd.b2 φB=BaseL
3 lsspropd.w φBW
4 lsspropd.p φxWyWx+Ky=x+Ly
5 lsspropd.s1 φxPyBxKyW
6 lsspropd.s2 φxPyBxKy=xLy
7 lsspropd.p1 φP=BaseScalarK
8 lsspropd.p2 φP=BaseScalarL
9 lsppropd.v1 φKX
10 lsppropd.v2 φLY
11 1 2 eqtr3d φBaseK=BaseL
12 11 pweqd φ𝒫BaseK=𝒫BaseL
13 1 2 3 4 5 6 7 8 lsspropd φLSubSpK=LSubSpL
14 13 rabeqdv φtLSubSpK|st=tLSubSpL|st
15 14 inteqd φtLSubSpK|st=tLSubSpL|st
16 12 15 mpteq12dv φs𝒫BaseKtLSubSpK|st=s𝒫BaseLtLSubSpL|st
17 eqid BaseK=BaseK
18 eqid LSubSpK=LSubSpK
19 eqid LSpanK=LSpanK
20 17 18 19 lspfval KXLSpanK=s𝒫BaseKtLSubSpK|st
21 9 20 syl φLSpanK=s𝒫BaseKtLSubSpK|st
22 eqid BaseL=BaseL
23 eqid LSubSpL=LSubSpL
24 eqid LSpanL=LSpanL
25 22 23 24 lspfval LYLSpanL=s𝒫BaseLtLSubSpL|st
26 10 25 syl φLSpanL=s𝒫BaseLtLSubSpL|st
27 16 21 26 3eqtr4d φLSpanK=LSpanL