Theorem bj-projeq 34299

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 487	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → 𝐵 = 𝐷)
2		simpl 485	. . . . . . 7 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → 𝐴 = 𝐶)
3	2	sneqd 4572	. . . . . 6 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝐴} = {𝐶})
4	1, 3	imaeq12d 5924	. . . . 5 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐵 “ {𝐴}) = (𝐷 “ {𝐶}))
5	4	eleq2d 2898	. . . 4 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ({𝑥} ∈ (𝐵 “ {𝐴}) ↔ {𝑥} ∈ (𝐷 “ {𝐶})))
6	5	abbidv 2885	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})} = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})})
7		df-bj-proj 34298	. . 3 ⊢ (𝐴 Proj 𝐵) = {𝑥 ∣ {𝑥} ∈ (𝐵 “ {𝐴})}
8		df-bj-proj 34298	. . 3 ⊢ (𝐶 Proj 𝐷) = {𝑥 ∣ {𝑥} ∈ (𝐷 “ {𝐶})}
9	6, 7, 8	3eqtr4g 2881	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷))
10	9	ex 415	1 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐷 → (𝐴 Proj 𝐵) = (𝐶 Proj 𝐷)))

Syntax hints:   → wi 4   ∧ wa 398   = wceq 1533   ∈ wcel 2110  {cab 2799  {csn 4560   “ cima 5552   Proj bj-cproj 34297

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4561  df-pr 4563  df-op 4567  df-br 5059  df-opab 5121  df-xp 5555  df-cnv 5557  df-dm 5559  df-rn 5560  df-res 5561  df-ima 5562  df-bj-proj 34298

This theorem is referenced by:  bj-projeq2  34300