Metamath Proof Explorer


Theorem relcnveq2

Description: Two ways of saying a relation is symmetric. (Contributed by Peter Mazsa, 28-Apr-2019)

Ref Expression
Assertion relcnveq2 ( Rel 𝑅 → ( ◡ 𝑅 = 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 ↔ 𝑦 𝑅 𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 cnvsym ⊢ ( ◡ 𝑅 ⊆ 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 → 𝑦 𝑅 𝑥 ) )
2 1 a1i ⊢ ( Rel 𝑅 → ( ◡ 𝑅 ⊆ 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 → 𝑦 𝑅 𝑥 ) ) )
3 dfrel2 ⊢ ( Rel 𝑅 ↔ ◡ ◡ 𝑅 = 𝑅 )
4 3 biimpi ⊢ ( Rel 𝑅 → ◡ ◡ 𝑅 = 𝑅 )
5 4 sseq1d ⊢ ( Rel 𝑅 → ( ◡ ◡ 𝑅 ⊆ ◡ 𝑅 ↔ 𝑅 ⊆ ◡ 𝑅 ) )
6 cnvsym ⊢ ( ◡ ◡ 𝑅 ⊆ ◡ 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 ◡ 𝑅 𝑦 → 𝑦 ◡ 𝑅 𝑥 ) )
7 5 6 bitr3di ⊢ ( Rel 𝑅 → ( 𝑅 ⊆ ◡ 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 ◡ 𝑅 𝑦 → 𝑦 ◡ 𝑅 𝑥 ) ) )
8 relbrcnvg ⊢ ( Rel 𝑅 → ( 𝑥 ◡ 𝑅 𝑦 ↔ 𝑦 𝑅 𝑥 ) )
9 relbrcnvg ⊢ ( Rel 𝑅 → ( 𝑦 ◡ 𝑅 𝑥 ↔ 𝑥 𝑅 𝑦 ) )
10 8 9 imbi12d ⊢ ( Rel 𝑅 → ( ( 𝑥 ◡ 𝑅 𝑦 → 𝑦 ◡ 𝑅 𝑥 ) ↔ ( 𝑦 𝑅 𝑥 → 𝑥 𝑅 𝑦 ) ) )
11 10 2albidv ⊢ ( Rel 𝑅 → ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 ◡ 𝑅 𝑦 → 𝑦 ◡ 𝑅 𝑥 ) ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑦 𝑅 𝑥 → 𝑥 𝑅 𝑦 ) ) )
12 7 11 bitrd ⊢ ( Rel 𝑅 → ( 𝑅 ⊆ ◡ 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑦 𝑅 𝑥 → 𝑥 𝑅 𝑦 ) ) )
13 2 12 anbi12d ⊢ ( Rel 𝑅 → ( ( ◡ 𝑅 ⊆ 𝑅 ∧ 𝑅 ⊆ ◡ 𝑅 ) ↔ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 → 𝑦 𝑅 𝑥 ) ∧ ∀ 𝑥 ∀ 𝑦 ( 𝑦 𝑅 𝑥 → 𝑥 𝑅 𝑦 ) ) ) )
14 eqss ⊢ ( ◡ 𝑅 = 𝑅 ↔ ( ◡ 𝑅 ⊆ 𝑅 ∧ 𝑅 ⊆ ◡ 𝑅 ) )
15 2albiim ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 ↔ 𝑦 𝑅 𝑥 ) ↔ ( ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 → 𝑦 𝑅 𝑥 ) ∧ ∀ 𝑥 ∀ 𝑦 ( 𝑦 𝑅 𝑥 → 𝑥 𝑅 𝑦 ) ) )
16 13 14 15 3bitr4g ⊢ ( Rel 𝑅 → ( ◡ 𝑅 = 𝑅 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑥 𝑅 𝑦 ↔ 𝑦 𝑅 𝑥 ) ) )