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Theorem 2reu5lem1 3446
Description: Lemma for 2reu5 3449. Note that ∃!𝑥𝐴∃!𝑦𝐵𝜑 does not mean "there is exactly one 𝑥 in 𝐴 and exactly one 𝑦 in 𝐵 such that 𝜑 holds;" see comment for 2eu5 2586. (Contributed by Alexander van der Vekens, 17-Jun-2017.)
Assertion
Ref Expression
2reu5lem1 (∃!𝑥𝐴 ∃!𝑦𝐵 𝜑 ↔ ∃!𝑥∃!𝑦(𝑥𝐴𝑦𝐵𝜑))
Distinct variable groups:   𝑦,𝐴   𝑥,𝐵   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝐴(𝑥)   𝐵(𝑦)

Proof of Theorem 2reu5lem1
StepHypRef Expression
1 df-reu 2948 . . 3 (∃!𝑦𝐵 𝜑 ↔ ∃!𝑦(𝑦𝐵𝜑))
21reubii 3158 . 2 (∃!𝑥𝐴 ∃!𝑦𝐵 𝜑 ↔ ∃!𝑥𝐴 ∃!𝑦(𝑦𝐵𝜑))
3 df-reu 2948 . . 3 (∃!𝑥𝐴 ∃!𝑦(𝑦𝐵𝜑) ↔ ∃!𝑥(𝑥𝐴 ∧ ∃!𝑦(𝑦𝐵𝜑)))
4 euanv 2563 . . . . . 6 (∃!𝑦(𝑥𝐴 ∧ (𝑦𝐵𝜑)) ↔ (𝑥𝐴 ∧ ∃!𝑦(𝑦𝐵𝜑)))
54bicomi 214 . . . . 5 ((𝑥𝐴 ∧ ∃!𝑦(𝑦𝐵𝜑)) ↔ ∃!𝑦(𝑥𝐴 ∧ (𝑦𝐵𝜑)))
6 3anass 1059 . . . . . . 7 ((𝑥𝐴𝑦𝐵𝜑) ↔ (𝑥𝐴 ∧ (𝑦𝐵𝜑)))
76bicomi 214 . . . . . 6 ((𝑥𝐴 ∧ (𝑦𝐵𝜑)) ↔ (𝑥𝐴𝑦𝐵𝜑))
87eubii 2520 . . . . 5 (∃!𝑦(𝑥𝐴 ∧ (𝑦𝐵𝜑)) ↔ ∃!𝑦(𝑥𝐴𝑦𝐵𝜑))
95, 8bitri 264 . . . 4 ((𝑥𝐴 ∧ ∃!𝑦(𝑦𝐵𝜑)) ↔ ∃!𝑦(𝑥𝐴𝑦𝐵𝜑))
109eubii 2520 . . 3 (∃!𝑥(𝑥𝐴 ∧ ∃!𝑦(𝑦𝐵𝜑)) ↔ ∃!𝑥∃!𝑦(𝑥𝐴𝑦𝐵𝜑))
113, 10bitri 264 . 2 (∃!𝑥𝐴 ∃!𝑦(𝑦𝐵𝜑) ↔ ∃!𝑥∃!𝑦(𝑥𝐴𝑦𝐵𝜑))
122, 11bitri 264 1 (∃!𝑥𝐴 ∃!𝑦𝐵 𝜑 ↔ ∃!𝑥∃!𝑦(𝑥𝐴𝑦𝐵𝜑))
Colors of variables: wff setvar class
Syntax hints:  wb 196  wa 383  w3a 1054  wcel 2030  ∃!weu 2498  ∃!wreu 2943
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-12 2087
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1056  df-tru 1526  df-ex 1745  df-nf 1750  df-eu 2502  df-reu 2948
This theorem is referenced by:  2reu5lem3  3448
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