Theorem frege55lem1c 44915

Description: Necessary deduction regarding substitution of value in equality. (Contributed by RP, 24-Dec-2019.)

Assertion

Ref	Expression
frege55lem1c	⊢ ((𝜑 → [𝐴 / 𝑥]𝑥 = 𝐵) → (𝜑 → 𝐴 = 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem frege55lem1c

Step	Hyp	Ref	Expression
1		df-sbc 3740	. . 3 ⊢ ([𝐴 / 𝑥]𝑥 = 𝐵 ↔ 𝐴 ∈ {𝑥 ∣ 𝑥 = 𝐵})
2		eqeq1 2765	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
3	2	elabg 3630	. . . 4 ⊢ (𝐴 ∈ {𝑥 ∣ 𝑥 = 𝐵} → (𝐴 ∈ {𝑥 ∣ 𝑥 = 𝐵} ↔ 𝐴 = 𝐵))
4	3	ibi 270	. . 3 ⊢ (𝐴 ∈ {𝑥 ∣ 𝑥 = 𝐵} → 𝐴 = 𝐵)
5	1, 4	sylbi 220	. 2 ⊢ ([𝐴 / 𝑥]𝑥 = 𝐵 → 𝐴 = 𝐵)
6	5	imim2i 17	1 ⊢ ((𝜑 → [𝐴 / 𝑥]𝑥 = 𝐵) → (𝜑 → 𝐴 = 𝐵))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2145  {cab 2739  [wsbc 3739

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-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-sbc 3740

This theorem is used by:  frege56c  44918