Theorem sbhypf 2905

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf 3172. (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 2863	. . 3 ⊢ y ∈ V
2		eqeq1 2359	. . 3 ⊢ (x = y → (x = A ↔ y = A))
3	1, 2	ceqsexv 2895	. 2 ⊢ (∃x(x = y ∧ x = A) ↔ y = A)
4		nfs1v 2106	. . . 4 ⊢ Ⅎx[y / x]φ
5		sbhypf.1	. . . 4 ⊢ Ⅎxψ
6	4, 5	nfbi 1834	. . 3 ⊢ Ⅎx([y / x]φ ↔ ψ)
7		sbequ12 1919	. . . . 5 ⊢ (x = y → (φ ↔ [y / x]φ))
8	7	bicomd 192	. . . 4 ⊢ (x = y → ([y / x]φ ↔ φ))
9		sbhypf.2	. . . 4 ⊢ (x = A → (φ ↔ ψ))
10	8, 9	sylan9bb 680	. . 3 ⊢ ((x = y ∧ x = A) → ([y / x]φ ↔ ψ))
11	6, 10	exlimi 1803	. 2 ⊢ (∃x(x = y ∧ x = A) → ([y / x]φ ↔ ψ))
12	3, 11	sylbir 204	1 ⊢ (y = A → ([y / x]φ ↔ ψ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∃wex 1541  Ⅎwnf 1544   = wceq 1642  [wsb 1648

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-v 2862

This theorem is used by:  mob2  3017  ralxpf  4828