Theorem bj-2upleq 37668

Description: Substitution property for ⦅ − , − ⦆. (Contributed by BJ, 6-Oct-2018.)

Assertion

Ref	Expression
bj-2upleq	⊢ (𝐴 = 𝐵 → (𝐶 = 𝐷 → ⦅𝐴, 𝐶⦆ = ⦅𝐵, 𝐷⦆))

Proof of Theorem bj-2upleq

Step	Hyp	Ref	Expression
1		bj-1upleq 37655	. . 3 ⊢ (𝐴 = 𝐵 → ⦅𝐴⦆ = ⦅𝐵⦆)
2		bj-xtageq 37644	. . 3 ⊢ (𝐶 = 𝐷 → ({1o} × tag 𝐶) = ({1o} × tag 𝐷))
3		uneq12 4117	. . . 4 ⊢ ((⦅𝐴⦆ = ⦅𝐵⦆ ∧ ({1o} × tag 𝐶) = ({1o} × tag 𝐷)) → (⦅𝐴⦆ ∪ ({1o} × tag 𝐶)) = (⦅𝐵⦆ ∪ ({1o} × tag 𝐷)))
4	3	ex 417	. . 3 ⊢ (⦅𝐴⦆ = ⦅𝐵⦆ → (({1o} × tag 𝐶) = ({1o} × tag 𝐷) → (⦅𝐴⦆ ∪ ({1o} × tag 𝐶)) = (⦅𝐵⦆ ∪ ({1o} × tag 𝐷))))
5	1, 2, 4	syl2im 41	. 2 ⊢ (𝐴 = 𝐵 → (𝐶 = 𝐷 → (⦅𝐴⦆ ∪ ({1o} × tag 𝐶)) = (⦅𝐵⦆ ∪ ({1o} × tag 𝐷))))
6		df-bj-2upl 37667	. . 3 ⊢ ⦅𝐴, 𝐶⦆ = (⦅𝐴⦆ ∪ ({1o} × tag 𝐶))
7		df-bj-2upl 37667	. . 3 ⊢ ⦅𝐵, 𝐷⦆ = (⦅𝐵⦆ ∪ ({1o} × tag 𝐷))
8	6, 7	eqeq12i 2781	. 2 ⊢ (⦅𝐴, 𝐶⦆ = ⦅𝐵, 𝐷⦆ ↔ (⦅𝐴⦆ ∪ ({1o} × tag 𝐶)) = (⦅𝐵⦆ ∪ ({1o} × tag 𝐷)))
9	5, 8	imbitrrdi 255	1 ⊢ (𝐴 = 𝐵 → (𝐶 = 𝐷 → ⦅𝐴, 𝐶⦆ = ⦅𝐵, 𝐷⦆))

Syntax hints:   → wi 4   = wceq 1570   ∪ cun 3903  {csn 4589   × cxp 5659  1_oc1o 8442  tag bj-ctag 37630  ⦅bj-c1upl 37653  ⦅bj-c2uple 37666

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rex 3090  df-v 3457  df-un 3910  df-opab 5174  df-xp 5667  df-bj-sngl 37622  df-bj-tag 37631  df-bj-1upl 37654  df-bj-2upl 37667

This theorem is referenced by:  bj-2uplth  37677