Theorem predisj 49890

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 7061	. . . . 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 6052	. . . . 5 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = (◡𝐹 “ ∅))
8		ima0 6075	. . . . 5 ⊢ (◡𝐹 “ ∅) = ∅
9	7, 8	eqtrdi 2812	. . . 4 ⊢ (𝜑 → (◡𝐹 “ (𝐴 ∩ 𝐵)) = ∅)
10	5, 9	eqtr3d 2798	. . 3 ⊢ (𝜑 → ((◡𝐹 “ 𝐴) ∩ (◡𝐹 “ 𝐵)) = ∅)
11	2, 10	ssdisjd 49887	. 2 ⊢ (𝜑 → (𝑆 ∩ (◡𝐹 “ 𝐵)) = ∅)
12	1, 11	ssdisjdr 49888	1 ⊢ (𝜑 → (𝑆 ∩ 𝑇) = ∅)

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279  ^◡ccnv 5650   “ cima 5654  Fun wfun 6531

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-pr 5391

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 2100  df-mo 2565  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-rn 5662  df-res 5663  df-ima 5664  df-fun 6539

This theorem is used by:  sepfsepc  50005