Metamath Proof Explorer


Theorem etransclem12

Description: C applied to N . (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses etransclem12.1 C = n 0 c 0 n 0 M | j = 0 M c j = n
etransclem12.2 φ N 0
Assertion etransclem12 φ C N = c 0 N 0 M | j = 0 M c j = N

Proof

Step Hyp Ref Expression
1 etransclem12.1 C = n 0 c 0 n 0 M | j = 0 M c j = n
2 etransclem12.2 φ N 0
3 oveq2 n = N 0 n = 0 N
4 3 oveq1d n = N 0 n 0 M = 0 N 0 M
5 eqeq2 n = N j = 0 M c j = n j = 0 M c j = N
6 4 5 rabeqbidv n = N c 0 n 0 M | j = 0 M c j = n = c 0 N 0 M | j = 0 M c j = N
7 ovex 0 N 0 M V
8 7 rabex c 0 N 0 M | j = 0 M c j = N V
9 8 a1i φ c 0 N 0 M | j = 0 M c j = N V
10 1 6 2 9 fvmptd3 φ C N = c 0 N 0 M | j = 0 M c j = N