Theorem bj-substw 37391

Description: Weak form of the LHS of bj-substax12 37390 proved from the core axiom schemes. Compare ax12w 2171. (Contributed by BJ, 26-May-2024.) (Proof modification is discouraged.)

Hypothesis

Ref	Expression
bj-substw.is	⊢ (𝑥 = 𝑡 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
bj-substw	⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))

Distinct variable group:   𝜓,𝑥

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

Proof of Theorem bj-substw

Step	Hyp	Ref	Expression
1		bj-substw.is	. . . . 5 ⊢ (𝑥 = 𝑡 → (𝜑 ↔ 𝜓))
2	1	pm5.32i 585	. . . 4 ⊢ ((𝑥 = 𝑡 ∧ 𝜑) ↔ (𝑥 = 𝑡 ∧ 𝜓))
3	2	exbii 1881	. . 3 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑡 ∧ 𝜓))
4		19.41v 1982	. . 3 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜓) ↔ (∃𝑥 𝑥 = 𝑡 ∧ 𝜓))
5	3, 4	bitri 278	. 2 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ (∃𝑥 𝑥 = 𝑡 ∧ 𝜓))
6	1	biimprcd 253	. . 3 ⊢ (𝜓 → (𝑥 = 𝑡 → 𝜑))
7	6	alrimiv 1960	. 2 ⊢ (𝜓 → ∀𝑥(𝑥 = 𝑡 → 𝜑))
8	5, 7	simplbiim 514	1 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))

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

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813

This theorem is used by: (None)