Theorem sbjust 2098

Description: Justification theorem for df-sb 2100 proved from Tarski's FOL axiom schemes. (Contributed by BJ, 22-Jan-2023.)

Assertion

Ref	Expression
sbjust	⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑧(𝑧 = 𝑡 → ∀𝑥(𝑥 = 𝑧 → 𝜑)))

Distinct variable groups:   𝑥,𝑦,𝑧   𝑦,𝑡,𝑧   𝜑,𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥, 𝑡)

Proof of Theorem sbjust

Step	Hyp	Ref	Expression
1		equequ1 2058	. . 3 ⊢ (𝑦 = 𝑧 → (𝑦 = 𝑡 ↔ 𝑧 = 𝑡))
2		equequ2 2059	. . . . 5 ⊢ (𝑦 = 𝑧 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑧))
3	2	imbi1d 344	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑧 → 𝜑)))
4	3	albidv 1953	. . 3 ⊢ (𝑦 = 𝑧 → (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑧 → 𝜑)))
5	1, 4	imbi12d 347	. 2 ⊢ (𝑦 = 𝑧 → ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑧 = 𝑡 → ∀𝑥(𝑥 = 𝑧 → 𝜑))))
6	5	cbvalvw 2069	1 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑧(𝑧 = 𝑡 → ∀𝑥(𝑥 = 𝑧 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568

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

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

This theorem is used by:  dfsb  2101