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Theorem bdsepnfALT 8942
Description: Alternate proof of bdsepnf 8941, not using bdsepnft 8940. (Contributed by BJ, 5-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypotheses
Ref Expression
bdsepnf.nf 𝑏φ
bdsepnf.1 BOUNDED φ
Assertion
Ref Expression
bdsepnfALT 𝑏x(x 𝑏 ↔ (x 𝑎 φ))
Distinct variable group:   𝑎,𝑏,x
Allowed substitution hints:   φ(x,𝑎,𝑏)

Proof of Theorem bdsepnfALT
Dummy variable y is distinct from all other variables.
StepHypRef Expression
1 bdsepnf.1 . . 3 BOUNDED φ
21bdsep2 8939 . 2 yx(x y ↔ (x 𝑎 φ))
3 nfv 1418 . . . . 5 𝑏 x y
4 nfv 1418 . . . . . 6 𝑏 x 𝑎
5 bdsepnf.nf . . . . . 6 𝑏φ
64, 5nfan 1454 . . . . 5 𝑏(x 𝑎 φ)
73, 6nfbi 1478 . . . 4 𝑏(x y ↔ (x 𝑎 φ))
87nfal 1465 . . 3 𝑏x(x y ↔ (x 𝑎 φ))
9 nfv 1418 . . 3 yx(x 𝑏 ↔ (x 𝑎 φ))
10 elequ2 1598 . . . . 5 (y = 𝑏 → (x yx 𝑏))
1110bibi1d 222 . . . 4 (y = 𝑏 → ((x y ↔ (x 𝑎 φ)) ↔ (x 𝑏 ↔ (x 𝑎 φ))))
1211albidv 1702 . . 3 (y = 𝑏 → (x(x y ↔ (x 𝑎 φ)) ↔ x(x 𝑏 ↔ (x 𝑎 φ))))
138, 9, 12cbvex 1636 . 2 (yx(x y ↔ (x 𝑎 φ)) ↔ 𝑏x(x 𝑏 ↔ (x 𝑎 φ)))
142, 13mpbi 133 1 𝑏x(x 𝑏 ↔ (x 𝑎 φ))
Colors of variables: wff set class
Syntax hints:   wa 97  wb 98  wal 1240  wnf 1346  wex 1378  BOUNDED wbd 8866
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-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-4 1397  ax-14 1402  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-bdsep 8938
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-cleq 2030  df-clel 2033
This theorem is referenced by: (None)
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