Theorem eqsbc1 3784

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2886. (Contributed by Andrew Salmon, 29-Jun-2011.) Avoid ax-13 2401. (Revised by Wolf Lammen, 29-Apr-2023.)

Assertion

Ref	Expression
eqsbc1	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))

Distinct variable group:   𝑥,𝐵

Allowed substitution hints:   𝐴(𝑥)   𝑉(𝑥)

Proof of Theorem eqsbc1

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq 3740	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝐴 / 𝑥]𝑥 = 𝐵))
2		eqeq1 2764	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 = 𝐵 ↔ 𝐴 = 𝐵))
3		sbsbc 3742	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ [𝑦 / 𝑥]𝑥 = 𝐵)
4		eqsb1 2886	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
5	3, 4	bitr3i 280	. 2 ⊢ ([𝑦 / 𝑥]𝑥 = 𝐵 ↔ 𝑦 = 𝐵)
6	1, 2, 5	vtoclbg 3519	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 = 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  [wsb 2098   ∈ wcel 2144  [wsbc 3738

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-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3739

This theorem is used by:  eqsbc2  3801  fmptsnd  7178  fvmptnn04if  23176  snfil  24192  f1omptsnlem  38368  mptsnunlem  38370  topdifinffinlem  38379  relowlpssretop  38396  iotavalb  45522  onfrALTlem5  45633  eqsbc2VD  45930  onfrALTlem5VD  45975