Theorem up1st2nd2 50114

Description: Rewrite the universal property predicate with separated parts. (Contributed by Zhi Wang, 23-Oct-2025.)

Hypothesis

Ref	Expression
up1st2nd2.1	⊢ (𝜑 → 𝑋 ∈ (𝐹(𝐷 UP 𝐸)𝑊))

Assertion

Ref	Expression
up1st2nd2	⊢ (𝜑 → (1st ‘𝑋)(𝐹(𝐷 UP 𝐸)𝑊)(2nd ‘𝑋))

Proof of Theorem up1st2nd2

Step	Hyp	Ref	Expression
1		relup 50109	. 2 ⊢ Rel (𝐹(𝐷 UP 𝐸)𝑊)
2		up1st2nd2.1	. 2 ⊢ (𝜑 → 𝑋 ∈ (𝐹(𝐷 UP 𝐸)𝑊))
3		1st2ndbr 8039	. 2 ⊢ ((Rel (𝐹(𝐷 UP 𝐸)𝑊) ∧ 𝑋 ∈ (𝐹(𝐷 UP 𝐸)𝑊)) → (1st ‘𝑋)(𝐹(𝐷 UP 𝐸)𝑊)(2nd ‘𝑋))
4	1, 2, 3	sylancr 599	1 ⊢ (𝜑 → (1st ‘𝑋)(𝐹(𝐷 UP 𝐸)𝑊)(2nd ‘𝑋))

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2145   class class class wbr 5103  Rel wrel 5660  ‘cfv 6533  (class class class)co 7413  1^st c1st 7984  2^nd c2nd 7985   UP cup 50099

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-11 2194  ax-12 2213  ax-ext 2732  ax-rep 5232  ax-sep 5251  ax-nul 5263  ax-pow 5330  ax-pr 5398  ax-un 7736

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 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-reu 3366  df-rab 3413  df-v 3452  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-pw 4559  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-iun 4953  df-br 5104  df-opab 5168  df-mpt 5187  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-iota 6489  df-fun 6535  df-fn 6536  df-f 6537  df-f1 6538  df-fo 6539  df-f1o 6540  df-fv 6541  df-ov 7416  df-oprab 7417  df-mpo 7418  df-1st 7986  df-2nd 7987  df-func 17947  df-up 50100

This theorem is used by: (None)