Theorem bj-projeq 37515

Description: Substitution property for Proj. (Contributed by BJ, 6-Apr-2019.)

Assertion

Ref	Expression
bj-projeq	⊢ (𝐴 = 𝐶 → (𝐵 = 𝐷 → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷)))

Proof of Theorem bj-projeq

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		simpr 489	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → 𝐵 = 𝐷)
2		simpl 487	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → 𝐴 = 𝐶)
3	2	sneqd 4606	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴} = {𝐶})
4	1, 3	imaeq12d 6064	. . . . 5 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐵 “ {𝐴}) = (𝐷 “ {𝐶}))
5	4	eleq2d 2855	. . . 4 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ({𝑥} ∈ (𝐵 “ {𝐴}) ↔ {𝑥} ∈ (𝐷 “ {𝐶})))
6	5	abbidv 2835	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})} = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})})
7		df-bj-proj 37514	. . 3 ⊢ (𝐴 Proj 𝐵) = {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})}
8		df-bj-proj 37514	. . 3 ⊢ (𝐶 Proj 𝐷) = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})}
9	6, 7, 8	3eqtr4g 2829	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷))
10	9	ex 417	1 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐷 → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷)))

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1567   ∈ wcel 2149  {cab 2747  {csn 4594   “ cima 5665   Proj bj-cproj 37513

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5114  df-opab 5178  df-xp 5668  df-cnv 5670  df-dm 5672  df-rn 5673  df-res 5674  df-ima 5675  df-bj-proj 37514

This theorem is referenced by:  bj-projeq2  37516