Theorem predisj 50015

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 7064	. . . . 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 6051	. . . . 5 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = (◡𝐹 “ ∅))
8		ima0 6074	. . . . 5 ⊢ (◡𝐹 “ ∅) = ∅
9	7, 8	eqtrdi 2811	. . . 4 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ∅)
10	5, 9	eqtr3d 2797	. . 3 ⊢ (𝜑 → ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)) = ∅)
11	2, 10	ssdisjd 50012	. 2 ⊢ (𝜑 → (𝑆 ∩ (◡𝐹 “ 𝐵)) = ∅)
12	1, 11	ssdisjdr 50013	1 ⊢ (𝜑 → (𝑆 ∩ 𝑇) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ⊆ wss 3895   ∩ cin 4089  ∅c0 4278  ^◡ccnv 5649   “ cima 5653  Fun wfun 6533

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-8 2146  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-mo 2564  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-id 5545  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-fun 6541

This theorem is used by:  sepfsepc  50136