Theorem incomOLD 4092

Description: Obsolete version of incom 4091 as of 12-Dec-2023. Commutative law for intersection of classes. Exercise 7 of [TakeutiZaring] p. 17. (Contributed by NM, 21-Jun-1993.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
incomOLD	⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)

Proof of Theorem incomOLD

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ancom 464	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴))
2		elin 3859	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3		elin 3859	. . 3 ⊢ (𝑥 ∈ (𝐵 ∩ 𝐴) ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴))
4	1, 2, 3	3bitr4i 306	. 2 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ 𝑥 ∈ (𝐵 ∩ 𝐴))
5	4	eqriv 2735	1 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)

Syntax hints:   ∧ wa 399   = wceq 1542   ∈ wcel 2114   ∩ cin 3842

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-ex 1787  df-sb 2075  df-clab 2717  df-cleq 2730  df-clel 2811  df-v 3400  df-in 3850

This theorem is referenced by: (None)