Theorem a9evsep 3900

Description: Derive a weakened version of ax-i9 1463, where 𝑥 and 𝑦 must be distinct, from Separation ax-sep 3896 and Extensionality ax-ext 2063. The theorem ¬ ∀𝑥¬ 𝑥 = 𝑦 also holds (ax9vsep 3901), but in intuitionistic logic ∃𝑥𝑥 = 𝑦 is stronger. (Contributed by Jim Kingdon, 25-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
a9evsep	⊢ ∃𝑥 𝑥 = 𝑦

Distinct variable group:   𝑥,𝑦

Proof of Theorem a9evsep

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ax-sep 3896	. 2 ⊢ ∃𝑥∀𝑧(𝑧 ∈ 𝑥 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧)))
2		id 19	. . . . . . . 8 ⊢ (𝑧 = 𝑧 → 𝑧 = 𝑧)
3	2	biantru 296	. . . . . . 7 ⊢ (𝑧 ∈ 𝑦 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧)))
4	3	bibi2i 225	. . . . . 6 ⊢ ((𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) ↔ (𝑧 ∈ 𝑥 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧))))
5	4	biimpri 131	. . . . 5 ⊢ ((𝑧 ∈ 𝑥 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧))) → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
6	5	alimi 1384	. . . 4 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧))) → ∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦))
7		ax-ext 2063	. . . 4 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝑦) → 𝑥 = 𝑦)
8	6, 7	syl 14	. . 3 ⊢ (∀𝑧(𝑧 ∈ 𝑥 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧))) → 𝑥 = 𝑦)
9	8	eximi 1531	. 2 ⊢ (∃𝑥∀𝑧(𝑧 ∈ 𝑥 ↔ (𝑧 ∈ 𝑦 ∧ (𝑧 = 𝑧 → 𝑧 = 𝑧))) → ∃𝑥 𝑥 = 𝑦)
10	1, 9	ax-mp 7	1 ⊢ ∃𝑥 𝑥 = 𝑦

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103  ∀wal 1282   = wceq 1284  ∃wex 1421   ∈ wcel 1433

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-ial 1467  ax-ext 2063  ax-sep 3896

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  ax9vsep  3901