Theorem equsb3 2114

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 2109	1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧)

Syntax hints:   ↔ wb 207  [wsb 2073

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-sb 2074

This theorem is referenced by:  equsb1v  2116  mo3  2568  sb8eulem  2602  sb8iota  6459  mo5f  32583  ss-ax8  36460  mptsnunlem  37707  wl-equsb3  37934  wl-mo3t  37954  wl-sb8eut  37956  wl-sb8eutv  37957  frege55lem1b  44346  sbeqal1  44849  icheq  47944