Metamath Proof Explorer


Theorem finxpeq2

Description: Equality theorem for Cartesian exponentiation. (Contributed by ML, 19-Oct-2020)

Ref Expression
Assertion finxpeq2 M = N U ↑↑ M = U ↑↑ N

Proof

Step Hyp Ref Expression
1 eleq1 M = N M ω N ω
2 opeq1 M = N M y = N y
3 rdgeq2 M y = N y rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y
4 2 3 syl M = N rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y
5 id M = N M = N
6 4 5 fveq12d M = N rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y M = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y N
7 6 eqeq2d M = N = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y M = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y N
8 1 7 anbi12d M = N M ω = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y M N ω = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y N
9 8 abbidv M = N y | M ω = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y M = y | N ω = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y N
10 df-finxp U ↑↑ M = y | M ω = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x M y M
11 df-finxp U ↑↑ N = y | N ω = rec n ω , x V if n = 1 𝑜 x U if x V × U n 1 st x n x N y N
12 9 10 11 3eqtr4g M = N U ↑↑ M = U ↑↑ N