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Theorem esumeq12dvaf 33995
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 2214 . . . . 5 (𝜑 → ∀𝑘 𝐴 = 𝐵)
4 esumeq12dvaf.3 . . . . . . 7 ((𝜑𝑘𝐴) → 𝐶 = 𝐷)
54ex 412 . . . . . 6 (𝜑 → (𝑘𝐴𝐶 = 𝐷))
61, 5ralrimi 3263 . . . . 5 (𝜑 → ∀𝑘𝐴 𝐶 = 𝐷)
7 mpteq12f 5254 . . . . 5 ((∀𝑘 𝐴 = 𝐵 ∧ ∀𝑘𝐴 𝐶 = 𝐷) → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
83, 6, 7syl2anc 583 . . . 4 (𝜑 → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
98oveq2d 7464 . . 3 (𝜑 → ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
109unieqd 4944 . 2 (𝜑 ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
11 df-esum 33992 . 2 Σ*𝑘𝐴𝐶 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶))
12 df-esum 33992 . 2 Σ*𝑘𝐵𝐷 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷))
1310, 11, 123eqtr4g 2805 1 (𝜑 → Σ*𝑘𝐴𝐶 = Σ*𝑘𝐵𝐷)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395  wal 1535   = wceq 1537  wnf 1781  wcel 2108  wral 3067   cuni 4931  cmpt 5249  (class class class)co 7448  0cc0 11184  +∞cpnf 11321  [,]cicc 13410  s cress 17287  *𝑠cxrs 17560   tsums ctsu 24155  Σ*cesum 33991
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-12 2178  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-uni 4932  df-br 5167  df-opab 5229  df-mpt 5250  df-iota 6525  df-fv 6581  df-ov 7451  df-esum 33992
This theorem is referenced by:  esumeq12dva  33996  esumeq1d  33999  esumeq2d  34001  esumpinfval  34037  measvunilem0  34177
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