Theorem clos1eq1 5875

Description: Equality law for closure. (Contributed by SF, 11-Feb-2015.)

Assertion

Ref	Expression
clos1eq1	⊢ (S = T → Clos1 (S, R) = Clos1 (T, R))

Proof of Theorem clos1eq1

Dummy variable a is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sseq1 3293	. . . . 5 ⊢ (S = T → (S ⊆ a ↔ T ⊆ a))
2	1	anbi1d 685	. . . 4 ⊢ (S = T → ((S ⊆ a ∧ (R “ a) ⊆ a) ↔ (T ⊆ a ∧ (R “ a) ⊆ a)))
3	2	abbidv 2468	. . 3 ⊢ (S = T → {a ∣ (S ⊆ a ∧ (R “ a) ⊆ a)} = {a ∣ (T ⊆ a ∧ (R “ a) ⊆ a)})
4		inteq 3930	. . 3 ⊢ ({a ∣ (S ⊆ a ∧ (R “ a) ⊆ a)} = {a ∣ (T ⊆ a ∧ (R “ a) ⊆ a)} → ∩{a ∣ (S ⊆ a ∧ (R “ a) ⊆ a)} = ∩{a ∣ (T ⊆ a ∧ (R “ a) ⊆ a)})
5	3, 4	syl 15	. 2 ⊢ (S = T → ∩{a ∣ (S ⊆ a ∧ (R “ a) ⊆ a)} = ∩{a ∣ (T ⊆ a ∧ (R “ a) ⊆ a)})
6		df-clos1 5874	. 2 ⊢ Clos1 (S, R) = ∩{a ∣ (S ⊆ a ∧ (R “ a) ⊆ a)}
7		df-clos1 5874	. 2 ⊢ Clos1 (T, R) = ∩{a ∣ (T ⊆ a ∧ (R “ a) ⊆ a)}
8	5, 6, 7	3eqtr4g 2410	1 ⊢ (S = T → Clos1 (S, R) = Clos1 (T, R))

This proof depends on syntax axioms:   → wi 4   ∧ wa 358   = wceq 1642  {cab 2339   ⊆ wss 3258  ∩cint 3927   “ cima 4723   Clos1 cclos1 5873

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ral 2620  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260  df-int 3928  df-clos1 5874

This theorem is used by:  clos1exg  5878  clos1basesucg  5885  spacval  6283  nchoicelem11  6300  nchoicelem16  6305  freceq12  6312  frecxp  6315