Theorem bnj610 35145

Description: Pass from equality (𝑥 = 𝐴) to substitution ([𝐴 / 𝑥]) without the distinct variable condition on 𝐴, 𝑥. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)

Hypotheses

Ref	Expression
bnj610.1	⊢ 𝐴 ∈ V
bnj610.2	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
bnj610.3	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓′))
bnj610.4	⊢ (𝑦 = 𝐴 → (𝜓′ ↔ 𝜓))

Assertion

Ref	Expression
bnj610	⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝜓)

Distinct variable groups:   𝑦,𝐴   𝜑,𝑦   𝜓,𝑦   𝑥,𝜓′   𝑥,𝑦

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

Proof of Theorem bnj610

Step	Hyp	Ref	Expression
1		vex 3458	. . . 4 ⊢ 𝑦 ∈ V
2		bnj610.3	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓′))
3	1, 2	sbcie 3784	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓′)
4	3	sbcbii 3799	. 2 ⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑦]𝜓′)
5		sbccow 3766	. 2 ⊢ ([𝐴 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑)
6		bnj610.1	. . 3 ⊢ 𝐴 ∈ V
7		bnj610.4	. . 3 ⊢ (𝑦 = 𝐴 → (𝜓′ ↔ 𝜓))
8	6, 7	sbcie 3784	. 2 ⊢ ([𝐴 / 𝑦]𝜓′ ↔ 𝜓)
9	4, 5, 8	3bitr3i 304	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   ∈ wcel 2142  Vcvv 3454  [wsbc 3743

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3456  df-sbc 3744

This theorem is used by:  bnj611  35315  bnj1000  35338