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Theorem esumeq12dvaf 34366
Description: Equality deduction for extended sum. (Contributed by Thierry Arnoux, 26-Mar-2017.)
Hypotheses
Ref Expression
esumeq12dvaf.1 𝑘𝜑
esumeq12dvaf.2 (𝜑𝐴 = 𝐵)
esumeq12dvaf.3 ((𝜑𝑘𝐴) → 𝐶 = 𝐷)
Assertion
Ref Expression
esumeq12dvaf (𝜑 → Σ*𝑘𝐴𝐶 = Σ*𝑘𝐵𝐷)

Proof of Theorem esumeq12dvaf
StepHypRef Expression
1 esumeq12dvaf.1 . . . . . 6 𝑘𝜑
2 esumeq12dvaf.2 . . . . . 6 (𝜑𝐴 = 𝐵)
31, 2alrimi 2255 . . . . 5 (𝜑 → ∀𝑘 𝐴 = 𝐵)
4 esumeq12dvaf.3 . . . . . . 7 ((𝜑𝑘𝐴) → 𝐶 = 𝐷)
54ex 417 . . . . . 6 (𝜑 → (𝑘𝐴𝐶 = 𝐷))
61, 5ralrimi 3269 . . . . 5 (𝜑 → ∀𝑘𝐴 𝐶 = 𝐷)
7 mpteq12f 5198 . . . . 5 ((∀𝑘 𝐴 = 𝐵 ∧ ∀𝑘𝐴 𝐶 = 𝐷) → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
83, 6, 7syl2anc 595 . . . 4 (𝜑 → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
98oveq2d 7427 . . 3 (𝜑 → ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
109unieqd 4887 . 2 (𝜑 ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
11 df-esum 34363 . 2 Σ*𝑘𝐴𝐶 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶))
12 df-esum 34363 . 2 Σ*𝑘𝐵𝐷 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷))
1310, 11, 123eqtr4g 2829 1 (𝜑 → Σ*𝑘𝐴𝐶 = Σ*𝑘𝐵𝐷)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 400  wal 1565   = wceq 1567  wnf 1810  wcel 2149  wral 3085   cuni 4874  cmpt 5194  (class class class)co 7411  0cc0 11100  +∞cpnf 11240  [,]cicc 13375  s cress 17290  *𝑠cxrs 17554   tsums ctsu 24252  Σ*cesum 34362
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-10 2182  ax-12 2219  ax-ext 2741
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-nf 1811  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rab 3423  df-v 3463  df-dif 3914  df-un 3916  df-ss 3928  df-nul 4293  df-if 4491  df-sn 4593  df-pr 4595  df-op 4599  df-uni 4875  df-br 5112  df-opab 5176  df-mpt 5195  df-iota 6493  df-fv 6545  df-ov 7414  df-esum 34363
This theorem is referenced by:  esumeq12dva  34367  esumeq1d  34370  esumeq2d  34372  esumpinfval  34408  measvunilem0  34548
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