Theorem sbhypf 3509

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf 3874. (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 2110	. 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 1569  Ⅎwnf 1815  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-sb 2099  df-cleq 2752

This theorem is used by:  mob2  3672  reu2eqd  3693  cbvrabcsfw  3887  cbvopab1  5178  cbvmptf  5204  ralxpf  5823  cbviotaw  6501  cbvriotaw  7386  tfisi  7870  ac6sf  10568  nn0ind-raph  12800  ac6sf2  33259  nn0min  33455  ac6gf  38742  fdc1  38756