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Theorem esumeq12dvaf 34227
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 2227 . . . . 5 (𝜑 → ∀𝑘 𝐴 = 𝐵)
4 esumeq12dvaf.3 . . . . . . 7 ((𝜑𝑘𝐴) → 𝐶 = 𝐷)
54ex 414 . . . . . 6 (𝜑 → (𝑘𝐴𝐶 = 𝐷))
61, 5ralrimi 3239 . . . . 5 (𝜑 → ∀𝑘𝐴 𝐶 = 𝐷)
7 mpteq12f 5160 . . . . 5 ((∀𝑘 𝐴 = 𝐵 ∧ ∀𝑘𝐴 𝐶 = 𝐷) → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
83, 6, 7syl2anc 591 . . . 4 (𝜑 → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
98oveq2d 7376 . . 3 (𝜑 → ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
109unieqd 4854 . 2 (𝜑 ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
11 df-esum 34224 . 2 Σ*𝑘𝐴𝐶 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶))
12 df-esum 34224 . 2 Σ*𝑘𝐵𝐷 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷))
1310, 11, 123eqtr4g 2801 1 (𝜑 → Σ*𝑘𝐴𝐶 = Σ*𝑘𝐵𝐷)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 397  wal 1546   = wceq 1548  wnf 1791  wcel 2121  wral 3055   cuni 4841  cmpt 5156  (class class class)co 7360  0cc0 11033  +∞cpnf 11171  [,]cicc 13296  s cress 17195  *𝑠cxrs 17459   tsums ctsu 24113  Σ*cesum 34223
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123  ax-9 2131  ax-10 2154  ax-12 2191  ax-ext 2713
This theorem depends on definitions:  df-bi 209  df-an 398  df-or 855  df-3an 1095  df-tru 1551  df-fal 1561  df-ex 1788  df-nf 1792  df-sb 2075  df-clab 2720  df-cleq 2733  df-clel 2816  df-ral 3056  df-rab 3394  df-v 3435  df-dif 3888  df-un 3890  df-ss 3902  df-nul 4265  df-if 4458  df-sn 4559  df-pr 4561  df-op 4565  df-uni 4842  df-br 5076  df-opab 5138  df-mpt 5157  df-iota 6445  df-fv 6497  df-ov 7363  df-esum 34224
This theorem is referenced by:  esumeq12dva  34228  esumeq1d  34231  esumeq2d  34233  esumpinfval  34269  measvunilem0  34409
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