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Theorem bj-ssbeq 34118
 Description: Substitution in an equality, disjoint variables case. Uses only ax-1 6 through ax-6 1970. It might be shorter to prove the result about composition of two substitutions and prove bj-ssbeq 34118 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 = 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-sb 2070 . 2 ([𝑡 / 𝑥]𝑦 = 𝑧 ↔ ∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)))
2 19.23v 1943 . . . . . 6 (∀𝑥(𝑥 = 𝑢𝑦 = 𝑧) ↔ (∃𝑥 𝑥 = 𝑢𝑦 = 𝑧))
3 ax6ev 1972 . . . . . . . 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 212 . . . . . 6 ((∃𝑥 𝑥 = 𝑢𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
82, 7bitri 278 . . . . 5 (∀𝑥(𝑥 = 𝑢𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
98imbi2i 339 . . . 4 ((𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)) ↔ (𝑢 = 𝑡𝑦 = 𝑧))
109albii 1821 . . 3 (∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)) ↔ ∀𝑢(𝑢 = 𝑡𝑦 = 𝑧))
11 19.23v 1943 . . . 4 (∀𝑢(𝑢 = 𝑡𝑦 = 𝑧) ↔ (∃𝑢 𝑢 = 𝑡𝑦 = 𝑧))
12 ax6ev 1972 . . . . . 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 212 . . . 4 ((∃𝑢 𝑢 = 𝑡𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
1711, 16bitri 278 . . 3 (∀𝑢(𝑢 = 𝑡𝑦 = 𝑧) ↔ 𝑦 = 𝑧)
1810, 17bitri 278 . 2 (∀𝑢(𝑢 = 𝑡 → ∀𝑥(𝑥 = 𝑢𝑦 = 𝑧)) ↔ 𝑦 = 𝑧)
191, 18bitri 278 1 ([𝑡 / 𝑥]𝑦 = 𝑧𝑦 = 𝑧)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209  ∀wal 1536  ∃wex 1781  [wsb 2069 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970 This theorem depends on definitions:  df-bi 210  df-ex 1782  df-sb 2070 This theorem is referenced by: (None)
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