Theorem cleq1 15010

Description: Equality of relations implies equality of closures. (Contributed by RP, 9-May-2020.)

Assertion

Ref	Expression
cleq1	⊢ (𝑅 = 𝑆 → ∩ {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = ∩ {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})

Distinct variable groups:   𝑅,𝑟   𝑆,𝑟

Allowed substitution hint:   𝜑(𝑟)

Proof of Theorem cleq1

Step	Hyp	Ref	Expression
1		cleq1lem 15009	. . 3 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝑟 ∧ 𝜑) ↔ (𝑆 ⊆ 𝑟 ∧ 𝜑)))
2	1	abbidv 2831	. 2 ⊢ (𝑅 = 𝑆 → {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})
3	2	inteqd 4913	1 ⊢ (𝑅 = 𝑆 → ∩ {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = ∩ {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1563  {cab 2743   ⊆ wss 3907  ∩ cint 4908

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-9 2155  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-ral 3080  df-rex 3090  df-ss 3924  df-int 4909

This theorem is referenced by:  trcleq1  15016  dfttc3gw  36896