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Theorem sbequ2 2287
Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2100. (Revised by BJ, 22-Dec-2020.) (Proof shortened by Wolf Lammen, 3-Feb-2024.)
Assertion
Ref Expression
sbequ2 (𝑥 = 𝑡 → ([𝑡 / 𝑥]𝜑𝜑))

Proof of Theorem sbequ2
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 dfsb 2101 . . . 4 ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
21biimpi 219 . . 3 ([𝑡 / 𝑥]𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
3 equvinva 2063 . . 3 (𝑥 = 𝑡 → ∃𝑦(𝑥 = 𝑦𝑡 = 𝑦))
4 equcomi 2050 . . . . . 6 (𝑡 = 𝑦𝑦 = 𝑡)
5 sp 2222 . . . . . 6 (∀𝑥(𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜑))
64, 5imim12i 63 . . . . 5 ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) → (𝑡 = 𝑦 → (𝑥 = 𝑦𝜑)))
76impcomd 417 . . . 4 ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) → ((𝑥 = 𝑦𝑡 = 𝑦) → 𝜑))
87aleximi 1865 . . 3 (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) → (∃𝑦(𝑥 = 𝑦𝑡 = 𝑦) → ∃𝑦𝜑))
92, 3, 8syl2im 41 . 2 ([𝑡 / 𝑥]𝜑 → (𝑥 = 𝑡 → ∃𝑦𝜑))
10 ax5e 1945 . 2 (∃𝑦𝜑𝜑)
119, 10syl6com 38 1 (𝑥 = 𝑡 → ([𝑡 / 𝑥]𝜑𝜑))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401  wal 1568  wex 1812  [wsb 2099
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2216
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100
This theorem is used by:  stdpc7  2288  sbequ12  2289  sb4a  2514  dfsb1  2515  dfsb2  2527  bj-ssbid2  37317  2pm13.193  45294  2pm13.193VD  45644
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