Theorem predisj 50246

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 7093	. . . . 5 ⊢ (Fun 𝐹 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)))
5	3, 4	syl 18	. . . 4 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)))
6		predisj.2	. . . . . 6 ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)
7	6	imaeq2d 6068	. . . . 5 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = (◡𝐹 “ ∅))
8		ima0 6091	. . . . 5 ⊢ (◡𝐹 “ ∅) = ∅
9	7, 8	eqtrdi 2814	. . . 4 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ∅)
10	5, 9	eqtr3d 2800	. . 3 ⊢ (𝜑 → ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)) = ∅)
11	2, 10	ssdisjd 50243	. 2 ⊢ (𝜑 → (𝑆 ∩ (◡𝐹 “ 𝐵)) = ∅)
12	1, 11	ssdisjdr 50244	1 ⊢ (𝜑 → (𝑆 ∩ 𝑇) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3900   ∩ cin 4092  ∅c0 4281  ^◡ccnv 5655   “ cima 5659  Fun wfun 6558

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-12 2216  ax-ext 2735  ax-sep 5253  ax-pr 5396

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-mo 2567  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-id 5551  df-xp 5662  df-rel 5663  df-cnv 5664  df-co 5665  df-dm 5666  df-rn 5667  df-res 5668  df-ima 5669  df-fun 6566

This theorem is used by:  sepfsepc  50367