Theorem sbhypf 2603

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf . (Contributed by Raph Levien, 10-Apr-2004.)

Hypotheses

Ref	Expression
sbhypf.1	⊢ Ⅎ𝑥𝜓
sbhypf.2	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
sbhypf	⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓))

Distinct variable groups:   𝑥,𝐴   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)   𝐴(𝑦)

Proof of Theorem sbhypf

Step	Hyp	Ref	Expression
1		vex 2560	. . 3 ⊢ 𝑦 ∈ V
2		eqeq1 2046	. . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴))
3	1, 2	ceqsexv 2593	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) ↔ 𝑦 = 𝐴)
4		nfs1v 1815	. . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑
5		sbhypf.1	. . . 4 ⊢ Ⅎ𝑥𝜓
6	4, 5	nfbi 1481	. . 3 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7		sbequ12 1654	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑))
8	7	bicomd 129	. . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
9		sbhypf.2	. . . 4 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
10	8, 9	sylan9bb 435	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓))
11	6, 10	exlimi 1485	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝑥 = 𝐴) → ([𝑦 / 𝑥]𝜑 ↔ 𝜓))
12	3, 11	sylbir 125	1 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  Ⅎwnf 1349  ∃wex 1381  [wsb 1645

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 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-v 2559

This theorem is referenced by:  mob2  2721  tfisi  4310  ralxpf  4482  rexxpf  4483  nn0ind-raph  8355