Theorem equsb3lem 1865

Description: Lemma for equsb3 1866. (Contributed by NM, 4-Dec-2005.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)

Assertion

Ref	Expression
equsb3lem	⊢ ([𝑥 / 𝑦]𝑦 = 𝑧 ↔ 𝑥 = 𝑧)

Distinct variable groups:   𝑦,𝑧   𝑥,𝑦

Proof of Theorem equsb3lem

Step	Hyp	Ref	Expression
1		ax-17 1459	. 2 ⊢ (𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)
2		equequ1 1638	. 2 ⊢ (𝑦 = 𝑥 → (𝑦 = 𝑧 ↔ 𝑥 = 𝑧))
3	1, 2	sbieh 1713	1 ⊢ ([𝑥 / 𝑦]𝑦 = 𝑧 ↔ 𝑥 = 𝑧)

Syntax hints:   ↔ wb 103   = wceq 1284  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  equsb3  1866