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Theorem esumeq12dvaf 31350
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 2215 . . . . 5 (𝜑 → ∀𝑘 𝐴 = 𝐵)
4 esumeq12dvaf.3 . . . . . . 7 ((𝜑𝑘𝐴) → 𝐶 = 𝐷)
54ex 416 . . . . . 6 (𝜑 → (𝑘𝐴𝐶 = 𝐷))
61, 5ralrimi 3210 . . . . 5 (𝜑 → ∀𝑘𝐴 𝐶 = 𝐷)
7 mpteq12f 5135 . . . . 5 ((∀𝑘 𝐴 = 𝐵 ∧ ∀𝑘𝐴 𝐶 = 𝐷) → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
83, 6, 7syl2anc 587 . . . 4 (𝜑 → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
98oveq2d 7165 . . 3 (𝜑 → ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
109unieqd 4838 . 2 (𝜑 ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
11 df-esum 31347 . 2 Σ*𝑘𝐴𝐶 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶))
12 df-esum 31347 . 2 Σ*𝑘𝐵𝐷 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷))
1310, 11, 123eqtr4g 2884 1 (𝜑 → Σ*𝑘𝐴𝐶 = Σ*𝑘𝐵𝐷)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 399  wal 1536   = wceq 1538  wnf 1785  wcel 2115  wral 3133   cuni 4824  cmpt 5132  (class class class)co 7149  0cc0 10535  +∞cpnf 10670  [,]cicc 12738  s cress 16484  *𝑠cxrs 16773   tsums ctsu 22737  Σ*cesum 31346
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-12 2179  ax-ext 2796
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-ral 3138  df-v 3482  df-un 3924  df-in 3926  df-ss 3936  df-sn 4551  df-pr 4553  df-op 4557  df-uni 4825  df-br 5053  df-opab 5115  df-mpt 5133  df-iota 6302  df-fv 6351  df-ov 7152  df-esum 31347
This theorem is referenced by:  esumeq12dva  31351  esumeq1d  31354  esumeq2d  31356  esumpinfval  31392  measvunilem0  31532
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