Theorem predisj 48730

Description: Preimages of disjoint sets are disjoint. (Contributed by Zhi Wang, 9-Sep-2024.)

Hypotheses

Ref	Expression
predisj.1	⊢ (𝜑 → Fun 𝐹)
predisj.2	⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)
predisj.3	⊢ (𝜑 → 𝑆 ⊆ (◡𝐹 “ 𝐴))
predisj.4	⊢ (𝜑 → 𝑇 ⊆ (◡𝐹 “ 𝐵))

Assertion

Ref	Expression
predisj	⊢ (𝜑 → (𝑆 ∩ 𝑇) = ∅)

Proof of Theorem predisj

Step	Hyp	Ref	Expression
1		predisj.4	. 2 ⊢ (𝜑 → 𝑇 ⊆ (◡𝐹 “ 𝐵))
2		predisj.3	. . 3 ⊢ (𝜑 → 𝑆 ⊆ (◡𝐹 “ 𝐴))
3		predisj.1	. . . . 5 ⊢ (𝜑 → Fun 𝐹)
4		inpreima 7084	. . . . 5 ⊢ (Fun 𝐹 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)))
5	3, 4	syl 17	. . . 4 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)))
6		predisj.2	. . . . . 6 ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)
7	6	imaeq2d 6078	. . . . 5 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = (◡𝐹 “ ∅))
8		ima0 6095	. . . . 5 ⊢ (◡𝐹 “ ∅) = ∅
9	7, 8	eqtrdi 2793	. . . 4 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ∅)
10	5, 9	eqtr3d 2779	. . 3 ⊢ (𝜑 → ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)) = ∅)
11	2, 10	ssdisjd 48727	. 2 ⊢ (𝜑 → (𝑆 ∩ (◡𝐹 “ 𝐵)) = ∅)
12	1, 11	ssdisjdr 48728	1 ⊢ (𝜑 → (𝑆 ∩ 𝑇) = ∅)

Syntax hints:   → wi 4   = wceq 1540   ∩ cin 3950   ⊆ wss 3951  ∅c0 4333  ^◡ccnv 5684   “ cima 5688  Fun wfun 6555

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-id 5578  df-xp 5691  df-rel 5692  df-cnv 5693  df-co 5694  df-dm 5695  df-rn 5696  df-res 5697  df-ima 5698  df-fun 6563

This theorem is referenced by:  sepfsepc  48825