Theorem bj-ssbeq 36654

Description: Substitution in an equality, disjoint variables case. Uses only ax-1 6 through ax-6 1967. It might be shorter to prove the result about composition of two substitutions and prove bj-ssbeq 36654 first with a DV condition on 𝑥, 𝑡, and then in the general case. (Contributed by BJ, 22-Dec-2020.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-ssbeq	⊢ ([𝑡 / 𝑥]𝑦 = 𝑧 ↔ 𝑦 = 𝑧)

Distinct variable groups:   𝑥,𝑦   𝑥,𝑧

Proof of Theorem bj-ssbeq

Dummy variable 𝑢 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-sb 2065	. 2 ⊢ ([𝑡 / 𝑥]𝑦 = 𝑧 ↔ ∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢 → 𝑦 = 𝑧)))
2		19.23v 1942	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝑢 → 𝑦 = 𝑧) ↔ (∃𝑥 𝑥 = 𝑢 → 𝑦 = 𝑧))
3		ax6ev 1969	. . . . . . . 8 ⊢ ∃𝑥 𝑥 = 𝑢
4		pm2.27 42	. . . . . . . 8 ⊢ (∃𝑥 𝑥 = 𝑢 → ((∃𝑥 𝑥 = 𝑢 → 𝑦 = 𝑧) → 𝑦 = 𝑧))
5	3, 4	ax-mp 5	. . . . . . 7 ⊢ ((∃𝑥 𝑥 = 𝑢 → 𝑦 = 𝑧) → 𝑦 = 𝑧)
6		ax-1 6	. . . . . . 7 ⊢ (𝑦 = 𝑧 → (∃𝑥 𝑥 = 𝑢 → 𝑦 = 𝑧))
7	5, 6	impbii 209	. . . . . 6 ⊢ ((∃𝑥 𝑥 = 𝑢 → 𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
8	2, 7	bitri 275	. . . . 5 ⊢ (∀𝑥(𝑥 = 𝑢 → 𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
9	8	imbi2i 336	. . . 4 ⊢ ((𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢 → 𝑦 = 𝑧)) ↔ (𝑢 = 𝑡 → 𝑦 = 𝑧))
10	9	albii 1819	. . 3 ⊢ (∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢 → 𝑦 = 𝑧)) ↔ ∀𝑢(𝑢 = 𝑡 → 𝑦 = 𝑧))
11		19.23v 1942	. . . 4 ⊢ (∀𝑢(𝑢 = 𝑡 → 𝑦 = 𝑧) ↔ (∃𝑢 𝑢 = 𝑡 → 𝑦 = 𝑧))
12		ax6ev 1969	. . . . . 6 ⊢ ∃𝑢 𝑢 = 𝑡
13		pm2.27 42	. . . . . 6 ⊢ (∃𝑢 𝑢 = 𝑡 → ((∃𝑢 𝑢 = 𝑡 → 𝑦 = 𝑧) → 𝑦 = 𝑧))
14	12, 13	ax-mp 5	. . . . 5 ⊢ ((∃𝑢 𝑢 = 𝑡 → 𝑦 = 𝑧) → 𝑦 = 𝑧)
15		ax-1 6	. . . . 5 ⊢ (𝑦 = 𝑧 → (∃𝑢 𝑢 = 𝑡 → 𝑦 = 𝑧))
16	14, 15	impbii 209	. . . 4 ⊢ ((∃𝑢 𝑢 = 𝑡 → 𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
17	11, 16	bitri 275	. . 3 ⊢ (∀𝑢(𝑢 = 𝑡 → 𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
18	10, 17	bitri 275	. 2 ⊢ (∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢 → 𝑦 = 𝑧)) ↔ 𝑦 = 𝑧)
19	1, 18	bitri 275	1 ⊢ ([𝑡 / 𝑥]𝑦 = 𝑧 ↔ 𝑦 = 𝑧)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538  ∃wex 1779  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967

This theorem depends on definitions:  df-bi 207  df-ex 1780  df-sb 2065

This theorem is referenced by: (None)