Theorem bj-projeq 37057

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 4587	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴} = {𝐶})
4	1, 3	imaeq12d 6014	. . . . 5 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐵 “ {𝐴}) = (𝐷 “ {𝐶}))
5	4	eleq2d 2819	. . . 4 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ({𝑥} ∈ (𝐵 “ {𝐴}) ↔ {𝑥} ∈ (𝐷 “ {𝐶})))
6	5	abbidv 2799	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})} = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})})
7		df-bj-proj 37056	. . 3 ⊢ (𝐴 Proj 𝐵) = {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})}
8		df-bj-proj 37056	. . 3 ⊢ (𝐶 Proj 𝐷) = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})}
9	6, 7, 8	3eqtr4g 2793	. 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 1541   ∈ wcel 2113  {cab 2711  {csn 4575   “ cima 5622   Proj bj-cproj 37055

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-rab 3397  df-v 3439  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4283  df-if 4475  df-sn 4576  df-pr 4578  df-op 4582  df-br 5094  df-opab 5156  df-xp 5625  df-cnv 5627  df-dm 5629  df-rn 5630  df-res 5631  df-ima 5632  df-bj-proj 37056

This theorem is referenced by:  bj-projeq2  37058