Metamath Proof Explorer


Theorem 2reu5lem1

Description: Lemma for 2reu5 . Note that E! x e. A E! y e. B ph does not mean "there is exactly one x in A and exactly one y in B such that ph holds"; see comment for 2eu5 . (Contributed by Alexander van der Vekens, 17-Jun-2017)

Ref Expression
Assertion 2reu5lem1 ( ∃! 𝑥𝐴 ∃! 𝑦𝐵 𝜑 ↔ ∃! 𝑥 ∃! 𝑦 ( 𝑥𝐴𝑦𝐵𝜑 ) )

Proof

Step Hyp Ref Expression
1 df-reu ( ∃! 𝑦𝐵 𝜑 ↔ ∃! 𝑦 ( 𝑦𝐵𝜑 ) )
2 1 reubii ( ∃! 𝑥𝐴 ∃! 𝑦𝐵 𝜑 ↔ ∃! 𝑥𝐴 ∃! 𝑦 ( 𝑦𝐵𝜑 ) )
3 df-reu ( ∃! 𝑥𝐴 ∃! 𝑦 ( 𝑦𝐵𝜑 ) ↔ ∃! 𝑥 ( 𝑥𝐴 ∧ ∃! 𝑦 ( 𝑦𝐵𝜑 ) ) )
4 euanv ( ∃! 𝑦 ( 𝑥𝐴 ∧ ( 𝑦𝐵𝜑 ) ) ↔ ( 𝑥𝐴 ∧ ∃! 𝑦 ( 𝑦𝐵𝜑 ) ) )
5 4 bicomi ( ( 𝑥𝐴 ∧ ∃! 𝑦 ( 𝑦𝐵𝜑 ) ) ↔ ∃! 𝑦 ( 𝑥𝐴 ∧ ( 𝑦𝐵𝜑 ) ) )
6 3anass ( ( 𝑥𝐴𝑦𝐵𝜑 ) ↔ ( 𝑥𝐴 ∧ ( 𝑦𝐵𝜑 ) ) )
7 6 bicomi ( ( 𝑥𝐴 ∧ ( 𝑦𝐵𝜑 ) ) ↔ ( 𝑥𝐴𝑦𝐵𝜑 ) )
8 7 eubii ( ∃! 𝑦 ( 𝑥𝐴 ∧ ( 𝑦𝐵𝜑 ) ) ↔ ∃! 𝑦 ( 𝑥𝐴𝑦𝐵𝜑 ) )
9 5 8 bitri ( ( 𝑥𝐴 ∧ ∃! 𝑦 ( 𝑦𝐵𝜑 ) ) ↔ ∃! 𝑦 ( 𝑥𝐴𝑦𝐵𝜑 ) )
10 9 eubii ( ∃! 𝑥 ( 𝑥𝐴 ∧ ∃! 𝑦 ( 𝑦𝐵𝜑 ) ) ↔ ∃! 𝑥 ∃! 𝑦 ( 𝑥𝐴𝑦𝐵𝜑 ) )
11 3 10 bitri ( ∃! 𝑥𝐴 ∃! 𝑦 ( 𝑦𝐵𝜑 ) ↔ ∃! 𝑥 ∃! 𝑦 ( 𝑥𝐴𝑦𝐵𝜑 ) )
12 2 11 bitri ( ∃! 𝑥𝐴 ∃! 𝑦𝐵 𝜑 ↔ ∃! 𝑥 ∃! 𝑦 ( 𝑥𝐴𝑦𝐵𝜑 ) )