Theorem equsb3 2109

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 2032	. 2 ⊢ (𝑥 = 𝑤 → (𝑥 = 𝑧 ↔ 𝑤 = 𝑧))
2		equequ1 2032	. 2 ⊢ (𝑤 = 𝑦 → (𝑤 = 𝑧 ↔ 𝑦 = 𝑧))
3	1, 2	sbievw2 2107	1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧)

Syntax hints:   ↔ wb 208  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-sb 2070

This theorem is referenced by:  equsb3rOLD  2111  equsb1v  2112  mo3  2648  sb8eulem  2684  sb8iota  6327  mo5f  30255  mptsnunlem  34621  wl-equsb3  34794  wl-mo3t  34814  wl-sb8eut  34815  wl-dfrmosb  34855  frege55lem1b  40248  sbeqal1  40737  icheq  43627