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Theorem bj-ssb1 33002
Description: A simplified definition of substitution in case of disjoint variables. See bj-ssb1a 33001 for the backward implication, which does not require ax-11 2198 (note that here, the version of ax-11 2198 with disjoint setvar metavariables would suffice). Compare sb6 2285. (Contributed by BJ, 22-Dec-2020.)
Assertion
Ref Expression
bj-ssb1 ([𝑡/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑡𝜑))
Distinct variable group:   𝑥,𝑡
Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem bj-ssb1
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 19.21v 2034 . . 3 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
21albii 1914 . 2 (∀𝑦𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
3 19.23v 2037 . . . . 5 (∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)) ↔ (∃𝑦 𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
4 equequ2 2123 . . . . . . . 8 (𝑦 = 𝑡 → (𝑥 = 𝑦𝑥 = 𝑡))
54imbi1d 332 . . . . . . 7 (𝑦 = 𝑡 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑡𝜑)))
65pm5.74i 262 . . . . . 6 ((𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
76albii 1914 . . . . 5 (∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
8 ax6ev 2071 . . . . . 6 𝑦 𝑦 = 𝑡
98a1bi 353 . . . . 5 ((𝑥 = 𝑡𝜑) ↔ (∃𝑦 𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
103, 7, 93bitr4ri 295 . . . 4 ((𝑥 = 𝑡𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
1110albii 1914 . . 3 (∀𝑥(𝑥 = 𝑡𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
12 alcom 2201 . . 3 (∀𝑥𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑦𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
1311, 12bitri 266 . 2 (∀𝑥(𝑥 = 𝑡𝜑) ↔ ∀𝑦𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
14 df-ssb 32989 . 2 ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
152, 13, 143bitr4ri 295 1 ([𝑡/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑡𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 197  wal 1650  wex 1874  [wssb 32988
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2069  ax-7 2105  ax-11 2198
This theorem depends on definitions:  df-bi 198  df-an 385  df-ex 1875  df-ssb 32989
This theorem is referenced by:  bj-ax12ssb  33004  bj-ssbssblem  33017  bj-ssbcom3lem  33018
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