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Theorem esumeq12dvaf 34330
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 2250 . . . . 5 (𝜑 → ∀𝑘 𝐴 = 𝐵)
4 esumeq12dvaf.3 . . . . . . 7 ((𝜑𝑘𝐴) → 𝐶 = 𝐷)
54ex 416 . . . . . 6 (𝜑 → (𝑘𝐴𝐶 = 𝐷))
61, 5ralrimi 3262 . . . . 5 (𝜑 → ∀𝑘𝐴 𝐶 = 𝐷)
7 mpteq12f 5187 . . . . 5 ((∀𝑘 𝐴 = 𝐵 ∧ ∀𝑘𝐴 𝐶 = 𝐷) → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
83, 6, 7syl2anc 593 . . . 4 (𝜑 → (𝑘𝐴𝐶) = (𝑘𝐵𝐷))
98oveq2d 7414 . . 3 (𝜑 → ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
109unieqd 4880 . 2 (𝜑 ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶)) = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷)))
11 df-esum 34327 . 2 Σ*𝑘𝐴𝐶 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐴𝐶))
12 df-esum 34327 . 2 Σ*𝑘𝐵𝐷 = ((ℝ*𝑠s (0[,]+∞)) tsums (𝑘𝐵𝐷))
1310, 11, 123eqtr4g 2824 1 (𝜑 → Σ*𝑘𝐴𝐶 = Σ*𝑘𝐵𝐷)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 399  wal 1560   = wceq 1562  wnf 1805  wcel 2144  wral 3078   cuni 4867  cmpt 5183  (class class class)co 7398  0cc0 11075  +∞cpnf 11215  [,]cicc 13354  s cress 17268  *𝑠cxrs 17532   tsums ctsu 24188  Σ*cesum 34326
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-10 2177  ax-12 2214  ax-ext 2736
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1101  df-tru 1565  df-fal 1575  df-ex 1802  df-nf 1806  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-ral 3079  df-rab 3417  df-v 3458  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4288  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5103  df-opab 5165  df-mpt 5184  df-iota 6479  df-fv 6531  df-ov 7401  df-esum 34327
This theorem is referenced by:  esumeq12dva  34331  esumeq1d  34334  esumeq2d  34336  esumpinfval  34372  measvunilem0  34512
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