Theorem cleq1 15141

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 15140	. . 3 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝑟 ∧ 𝜑) ↔ (𝑆 ⊆ 𝑟 ∧ 𝜑)))
2	1	abbidv 2826	. 2 ⊢ (𝑅 = 𝑆 → {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})
3	2	inteqd 4911	1 ⊢ (𝑅 = 𝑆 → ∩ {𝑟 ∣ (𝑅 ⊆ 𝑟 ∧ 𝜑)} = ∩ {𝑟 ∣ (𝑆 ⊆ 𝑟 ∧ 𝜑)})

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569  {cab 2738   ⊆ wss 3895  ∩ cint 4906

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-ral 3077  df-rex 3087  df-ss 3896  df-int 4907

This theorem is used by:  trcleq1  15147  dfttc3gw  37403