Metamath Proof Explorer


Theorem esumeq12dvaf

Description: Equality deduction for extended sum. (Contributed by Thierry Arnoux, 26-Mar-2017)

Ref Expression
Hypotheses esumeq12dvaf.1 kφ
esumeq12dvaf.2 φA=B
esumeq12dvaf.3 φkAC=D
Assertion esumeq12dvaf φ*kAC=*kBD

Proof

Step Hyp Ref Expression
1 esumeq12dvaf.1 kφ
2 esumeq12dvaf.2 φA=B
3 esumeq12dvaf.3 φkAC=D
4 1 2 alrimi φkA=B
5 3 ex φkAC=D
6 1 5 ralrimi φkAC=D
7 mpteq12f kA=BkAC=DkAC=kBD
8 4 6 7 syl2anc φkAC=kBD
9 8 oveq2d φ𝑠*𝑠0+∞tsumskAC=𝑠*𝑠0+∞tsumskBD
10 9 unieqd φ𝑠*𝑠0+∞tsumskAC=𝑠*𝑠0+∞tsumskBD
11 df-esum *kAC=𝑠*𝑠0+∞tsumskAC
12 df-esum *kBD=𝑠*𝑠0+∞tsumskBD
13 10 11 12 3eqtr4g φ*kAC=*kBD