Theorem pgindlem 50990

Description: Lemma for pgind 50992. (Contributed by Emmett Weisz, 27-May-2024.) (New usage is discouraged.)

Assertion

Ref	Expression
pgindlem	⊢ (𝑥 ∈ (𝒫 𝑧 × 𝒫 𝑧) → ((1st ‘𝑥) ∪ (2nd ‘𝑥)) ⊆ 𝑧)

Proof of Theorem pgindlem

Step	Hyp	Ref	Expression
1		xp1st 8035	. . 3 ⊢ (𝑥 ∈ (𝒫 𝑧 × 𝒫 𝑧) → (1st ‘𝑥) ∈ 𝒫 𝑧)
2	1	elpwid 4565	. 2 ⊢ (𝑥 ∈ (𝒫 𝑧 × 𝒫 𝑧) → (1st ‘𝑥) ⊆ 𝑧)
3		xp2nd 8036	. . 3 ⊢ (𝑥 ∈ (𝒫 𝑧 × 𝒫 𝑧) → (2nd ‘𝑥) ∈ 𝒫 𝑧)
4	3	elpwid 4565	. 2 ⊢ (𝑥 ∈ (𝒫 𝑧 × 𝒫 𝑧) → (2nd ‘𝑥) ⊆ 𝑧)
5	2, 4	unssd 4189	1 ⊢ (𝑥 ∈ (𝒫 𝑧 × 𝒫 𝑧) → ((1st ‘𝑥) ∪ (2nd ‘𝑥)) ⊆ 𝑧)

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2144   ⊆ wss 3896   ∪ cun 4148  𝒫 cpw 4556   × cxp 5648  ‘cfv 6541  1^st c1st 8001  2^nd c2nd 8002

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-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-nul 5259  ax-pr 5390  ax-un 7753

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-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-pw 4558  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-br 5103  df-opab 5167  df-mpt 5186  df-id 5545  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-iota 6496  df-fun 6543  df-fv 6549  df-1st 8003  df-2nd 8004

This theorem is used by:  pgindnf  50991