Theorem equsb3 2108

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 2031	. 2 ⊢ (𝑥 = 𝑤 → (𝑥 = 𝑧 ↔ 𝑤 = 𝑧))
2		equequ1 2031	. 2 ⊢ (𝑤 = 𝑦 → (𝑤 = 𝑧 ↔ 𝑦 = 𝑧))
3	1, 2	sbievw2 2106	1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧)

Syntax hints:   ↔ wb 208  [wsb 2068

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1780  df-sb 2069

This theorem is referenced by:  equsb3rOLD  2110  equsb1v  2111  mo3  2647  sb8eulem  2683  sb8iota  6328  mo5f  30256  mptsnunlem  34623  wl-equsb3  34796  wl-mo3t  34816  wl-sb8eut  34817  wl-dfrmosb  34857  frege55lem1b  40247  sbeqal1  40736  icheq  43627