Theorem invdisjrabw 5138

Description: Version of invdisjrab 5139 with a disjoint variable condition, which does not require ax-13 2366. (Contributed by GG, 26-Jan-2024.)

Assertion

Ref	Expression
invdisjrabw	⊢ Disj 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦}

Distinct variable groups:   𝑥,𝐵   𝑦,𝐶   𝑥,𝑦

Allowed substitution hints:   𝐴(𝑥,𝑦)   𝐵(𝑦)   𝐶(𝑥)

Proof of Theorem invdisjrabw

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		nfcv 2892	. . . . 5 ⊢ Ⅎ𝑥𝑧
2		nfcv 2892	. . . . 5 ⊢ Ⅎ𝑥𝐵
3		nfcsb1v 3917	. . . . . 6 ⊢ Ⅎ𝑥⦋𝑧 / 𝑥⦌𝐶
4	3	nfeq1 2908	. . . . 5 ⊢ Ⅎ𝑥⦋𝑧 / 𝑥⦌𝐶 = 𝑦
5		csbeq1a 3906	. . . . . 6 ⊢ (𝑥 = 𝑧 → 𝐶 = ⦋𝑧 / 𝑥⦌𝐶)
6	5	eqeq1d 2728	. . . . 5 ⊢ (𝑥 = 𝑧 → (𝐶 = 𝑦 ↔ ⦋𝑧 / 𝑥⦌𝐶 = 𝑦))
7	1, 2, 4, 6	elrabf 3677	. . . 4 ⊢ (𝑧 ∈ {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦} ↔ (𝑧 ∈ 𝐵 ∧ ⦋𝑧 / 𝑥⦌𝐶 = 𝑦))
8		simprr 771	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∧ (𝑧 ∈ 𝐵 ∧ ⦋𝑧 / 𝑥⦌𝐶 = 𝑦)) → ⦋𝑧 / 𝑥⦌𝐶 = 𝑦)
9	7, 8	sylan2b 592	. . 3 ⊢ ((𝑦 ∈ 𝐴 ∧ 𝑧 ∈ {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦}) → ⦋𝑧 / 𝑥⦌𝐶 = 𝑦)
10	9	rgen2 3188	. 2 ⊢ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦}⦋𝑧 / 𝑥⦌𝐶 = 𝑦
11		invdisj 5137	. 2 ⊢ (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦}⦋𝑧 / 𝑥⦌𝐶 = 𝑦 → Disj 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦})
12	10, 11	ax-mp 5	1 ⊢ Disj 𝑦 ∈ 𝐴 {𝑥 ∈ 𝐵 ∣ 𝐶 = 𝑦}

Syntax hints:   ∧ wa 394   = wceq 1534   ∈ wcel 2099  ∀wral 3051  {crab 3419  ⦋csb 3892  Disj wdisj 5118

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1537  df-ex 1775  df-nf 1779  df-sb 2061  df-mo 2529  df-clab 2704  df-cleq 2718  df-clel 2803  df-nfc 2878  df-ral 3052  df-rmo 3364  df-rab 3420  df-v 3464  df-sbc 3777  df-csb 3893  df-disj 5119

This theorem is referenced by: (None)