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 2058	. 2 ⊢ (𝑥 = 𝑤 → (𝑥 = 𝑧 ↔ 𝑤 = 𝑧))
2		equequ1 2058	. 2 ⊢ (𝑤 = 𝑦 → (𝑤 = 𝑧 ↔ 𝑦 = 𝑧))
3	1, 2	sbievw2 2135	1 ⊢ ([𝑦 / 𝑥]𝑥 = 𝑧 ↔ 𝑦 = 𝑧)

This proof depends on syntax axioms:   ↔ wb 209  [wsb 2099

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

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100

This theorem is used by:  equsb1v  2142  mo3  2590  sb8eulem  2624  sb8iota  6498  mo5f  33067  ss-ax8  36984  mptsnunlem  38229  wl-equsb3  38456  wl-mo3t  38476  wl-sb8eut  38478  wl-sb8eutv  38479  frege55lem1b  44854  sbeqal1  45341  icheq  48488