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Theorem alexeq 2664
Description: Two ways to express substitution of A for x in φ. (Contributed by NM, 2-Mar-1995.)
Hypothesis
Ref Expression
alexeq.1 A V
Assertion
Ref Expression
alexeq (x(x = Aφ) ↔ x(x = A φ))
Distinct variable group:   x,A
Allowed substitution hint:   φ(x)

Proof of Theorem alexeq
Dummy variable y is distinct from all other variables.
StepHypRef Expression
1 alexeq.1 . . 3 A V
2 eqeq2 2046 . . . . 5 (y = A → (x = yx = A))
32anbi1d 438 . . . 4 (y = A → ((x = y φ) ↔ (x = A φ)))
43exbidv 1703 . . 3 (y = A → (x(x = y φ) ↔ x(x = A φ)))
52imbi1d 220 . . . 4 (y = A → ((x = yφ) ↔ (x = Aφ)))
65albidv 1702 . . 3 (y = A → (x(x = yφ) ↔ x(x = Aφ)))
7 sb56 1762 . . 3 (x(x = y φ) ↔ x(x = yφ))
81, 4, 6, 7vtoclb 2605 . 2 (x(x = A φ) ↔ x(x = Aφ))
98bicomi 123 1 (x(x = Aφ) ↔ x(x = A φ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98  wal 1240   = wceq 1242  wex 1378   wcel 1390  Vcvv 2551
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-v 2553
This theorem is referenced by:  ceqex  2665
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