Theorem equsb3 2140

Description: Substitution in an equality. (Contributed by Raph Levien and FL, 4-Dec-2005.) Reduce axiom usage. (Revised by Wolf Lammen, 23-Jul-2023.)

Assertion

Ref	Expression
equsb3	⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧)

Distinct variable group:   𝑥,𝑧

Proof of Theorem equsb3

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ1 2048	. 2 ⊢ (𝑥 = 𝑤 → (𝑥 = 𝑧 ↔ 𝑤 = 𝑧))
2		equequ1 2048	. 2 ⊢ (𝑤 = 𝑦 → (𝑤 = 𝑧 ↔ 𝑦 = 𝑧))
3	1, 2	sbievw2 2135	1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧)

Syntax hints:   ↔ wb 209  [wsb 2093

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1803  df-sb 2094

This theorem is referenced by:  equsb1v  2142  mo3  2594  sb8eulem  2628  sb8iota  6492  mo5f  32741  ss-ax8  36593  mptsnunlem  37839  wl-equsb3  38066  wl-mo3t  38086  wl-sb8eut  38088  wl-sb8eutv  38089  frege55lem1b  44478  sbeqal1  44967  icheq  48067