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Theorem bj-ssbeq 32925
Description: Substitution in an equality, disjoint variables case. Uses only ax-1--6. It might be shorter to prove the result about composition of two substitutions and prove bj-ssbeq 32925 first with a DV on x,t, and then in the general case. (Contributed by BJ, 22-Dec-2020.)
Assertion
Ref Expression
bj-ssbeq ([𝑡/𝑥]b𝑦 = 𝑧𝑦 = 𝑧)
Distinct variable groups:   𝑥,𝑦   𝑥,𝑧

Proof of Theorem bj-ssbeq
Dummy variable 𝑢 is distinct from all other variables.
StepHypRef Expression
1 df-ssb 32918 . 2 ([𝑡/𝑥]b𝑦 = 𝑧 ↔ ∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)))
2 19.23v 2012 . . . . . 6 (∀𝑥(𝑥 = 𝑢𝑦 = 𝑧) ↔ (∃𝑥 𝑥 = 𝑢𝑦 = 𝑧))
3 ax6ev 2048 . . . . . . . 8 𝑥 𝑥 = 𝑢
4 pm2.27 42 . . . . . . . 8 (∃𝑥 𝑥 = 𝑢 → ((∃𝑥 𝑥 = 𝑢𝑦 = 𝑧) → 𝑦 = 𝑧))
53, 4ax-mp 5 . . . . . . 7 ((∃𝑥 𝑥 = 𝑢𝑦 = 𝑧) → 𝑦 = 𝑧)
6 ax-1 6 . . . . . . 7 (𝑦 = 𝑧 → (∃𝑥 𝑥 = 𝑢𝑦 = 𝑧))
75, 6impbii 199 . . . . . 6 ((∃𝑥 𝑥 = 𝑢𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
82, 7bitri 264 . . . . 5 (∀𝑥(𝑥 = 𝑢𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
98imbi2i 325 . . . 4 ((𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)) ↔ (𝑢 = 𝑡𝑦 = 𝑧))
109albii 1888 . . 3 (∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)) ↔ ∀𝑢(𝑢 = 𝑡𝑦 = 𝑧))
11 19.23v 2012 . . . 4 (∀𝑢(𝑢 = 𝑡𝑦 = 𝑧) ↔ (∃𝑢 𝑢 = 𝑡𝑦 = 𝑧))
12 ax6ev 2048 . . . . . 6 𝑢 𝑢 = 𝑡
13 pm2.27 42 . . . . . 6 (∃𝑢 𝑢 = 𝑡 → ((∃𝑢 𝑢 = 𝑡𝑦 = 𝑧) → 𝑦 = 𝑧))
1412, 13ax-mp 5 . . . . 5 ((∃𝑢 𝑢 = 𝑡𝑦 = 𝑧) → 𝑦 = 𝑧)
15 ax-1 6 . . . . 5 (𝑦 = 𝑧 → (∃𝑢 𝑢 = 𝑡𝑦 = 𝑧))
1614, 15impbii 199 . . . 4 ((∃𝑢 𝑢 = 𝑡𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
1711, 16bitri 264 . . 3 (∀𝑢(𝑢 = 𝑡𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
1810, 17bitri 264 . 2 (∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)) ↔ 𝑦 = 𝑧)
191, 18bitri 264 1 ([𝑡/𝑥]b𝑦 = 𝑧𝑦 = 𝑧)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wal 1622  wex 1845  [wssb 32917
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1863  ax-4 1878  ax-5 1980  ax-6 2046
This theorem depends on definitions:  df-bi 197  df-ex 1846  df-ssb 32918
This theorem is referenced by: (None)
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