Theorem cleq1 15025

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 15024	. . 3 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝑟 ∧ 𝜑) ↔ (𝑆 ⊆ 𝑟 ∧ 𝜑)))
2	1	abbidv 2829	. 2 ⊢ (𝑅 = 𝑆 → {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})
3	2	inteqd 4917	1 ⊢ (𝑅 = 𝑆 → ∩ {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = ∩ {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})

This proof depends on syntax axioms:   → wi 4   ∧ wa 400   = wceq 1570  {cab 2741   ⊆ wss 3905  ∩ cint 4912

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-9 2153  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-ral 3080  df-rex 3090  df-ss 3922  df-int 4913

This theorem is used by:  trcleq1  15031  dfttc3gw  37062