Theorem bj-equsexval 37323

Description: Special case of equsexv 2307 proved from core axioms, ax-10 2179 (modal5), and hba1 2331 (modal4). (Contributed by BJ, 29-Dec-2020.) (Proof modification is discouraged.)

Hypothesis

Ref	Expression
bj-equsexval.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ ∀𝑥𝜓))

Assertion

Ref	Expression
bj-equsexval	⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥𝜓)

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑥, 𝑦)

Proof of Theorem bj-equsexval

Step	Hyp	Ref	Expression
1		bj-equsexval.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ ∀𝑥𝜓))
2	1	pm5.32i 585	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝑦 ∧ ∀𝑥𝜓))
3	2	exbii 1881	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ ∀𝑥𝜓))
4		ax6ev 2002	. . 3 ⊢ ∃𝑥 𝑥 = 𝑦
5		bj-19.41al 37322	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ ∀𝑥𝜓) ↔ (∃𝑥 𝑥 = 𝑦 ∧ ∀𝑥𝜓))
6	4, 5	mpbiran 722	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ ∀𝑥𝜓) ↔ ∀𝑥𝜓)
7	3, 6	bitri 278	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817

This theorem is used by: (None)