Theorem bj-projeq 36973

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 4597	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴} = {𝐶})
4	1, 3	imaeq12d 6021	. . . . 5 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐵 “ {𝐴}) = (𝐷 “ {𝐶}))
5	4	eleq2d 2814	. . . 4 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ({𝑥} ∈ (𝐵 “ {𝐴}) ↔ {𝑥} ∈ (𝐷 “ {𝐶})))
6	5	abbidv 2795	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})} = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})})
7		df-bj-proj 36972	. . 3 ⊢ (𝐴 Proj 𝐵) = {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})}
8		df-bj-proj 36972	. . 3 ⊢ (𝐶 Proj 𝐷) = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})}
9	6, 7, 8	3eqtr4g 2789	. 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 1540   ∈ wcel 2109  {cab 2707  {csn 4585   “ cima 5634   Proj bj-cproj 36971

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-rab 3403  df-v 3446  df-dif 3914  df-un 3916  df-in 3918  df-ss 3928  df-nul 4293  df-if 4485  df-sn 4586  df-pr 4588  df-op 4592  df-br 5103  df-opab 5165  df-xp 5637  df-cnv 5639  df-dm 5641  df-rn 5642  df-res 5643  df-ima 5644  df-bj-proj 36972

This theorem is referenced by:  bj-projeq2  36974