Theorem bj-projeq 37318

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 484	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → 𝐵 = 𝐷)
2		simpl 482	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → 𝐴 = 𝐶)
3	2	sneqd 4580	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴} = {𝐶})
4	1, 3	imaeq12d 6021	. . . . 5 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐵 “ {𝐴}) = (𝐷 “ {𝐶}))
5	4	eleq2d 2823	. . . 4 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ({𝑥} ∈ (𝐵 “ {𝐴}) ↔ {𝑥} ∈ (𝐷 “ {𝐶})))
6	5	abbidv 2803	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})} = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})})
7		df-bj-proj 37317	. . 3 ⊢ (𝐴 Proj 𝐵) = {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})}
8		df-bj-proj 37317	. . 3 ⊢ (𝐶 Proj 𝐷) = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})}
9	6, 7, 8	3eqtr4g 2797	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷))
10	9	ex 412	1 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐷 → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷)))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1542   ∈ wcel 2114  {cab 2715  {csn 4568   “ cima 5628   Proj bj-cproj 37316

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-br 5087  df-opab 5149  df-xp 5631  df-cnv 5633  df-dm 5635  df-rn 5636  df-res 5637  df-ima 5638  df-bj-proj 37317

This theorem is referenced by:  bj-projeq2  37319