Theorem bj-ssblem1 31819

Description: A lemma for the definiens of df-sb 1868. (Contributed by BJ, 22-Dec-2020.)

Assertion

Ref	Expression
bj-ssblem1	⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))

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

Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem bj-ssblem1

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ1 1939	. . 3 ⊢ (𝑦 = 𝑧 → (𝑦 = 𝑡 ↔ 𝑧 = 𝑡))
2		equequ2 1940	. . . . 5 ⊢ (𝑦 = 𝑧 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑧))
3	2	imbi1d 330	. . . 4 ⊢ (𝑦 = 𝑧 → ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑧 → 𝜑)))
4	3	albidv 1836	. . 3 ⊢ (𝑦 = 𝑧 → (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑧 → 𝜑)))
5	1, 4	imbi12d 333	. 2 ⊢ (𝑦 = 𝑧 → ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑧 = 𝑡 → ∀𝑥(𝑥 = 𝑧 → 𝜑))))
6	5	spw 1954	1 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))

Syntax hints:   → wi 4  ∀wal 1473

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696

This theorem is referenced by: (None)