Theorem sbhypf 3509

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf 3875. (Contributed by Raph Levien, 10-Apr-2004.) (Proof shortened by Wolf Lammen, 25-Jan-2025.)

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		sbhypf.2	. . 3 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
2	1	sbimi 2111	. 2 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐴 → [𝑦 / 𝑥](𝜑 ↔ 𝜓))
3		eqsb1 2886	. 2 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐴 ↔ 𝑦 = 𝐴)
4		sbhypf.1	. . . 4 ⊢ Ⅎ𝑥𝜓
5	4	sbf 2304	. . 3 ⊢ ([𝑦 / 𝑥]𝜓 ↔ 𝜓)
6	5	sblbis 2341	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ 𝜓))
7	2, 3, 6	3imtr3i 294	1 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570  Ⅎwnf 1816  [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-9 2155  ax-10 2178  ax-12 2213  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2100  df-cleq 2752

This theorem is used by:  mob2  3673  reu2eqd  3694  cbvrabcsfw  3888  cbvopab1  5179  cbvmptf  5205  ralxpf  5826  cbviotaw  6496  cbvriotaw  7380  tfisi  7856  ac6sf  10492  nn0ind-raph  12722  ac6sf2  33096  nn0min  33292  ac6gf  38483  fdc1  38497