Theorem sbeqalbi 38601

Description: When both 𝑥 and 𝑧 and 𝑦 and 𝑧 are both distinct, then the converse of sbeqal1 holds as well. (Contributed by Andrew Salmon, 2-Jun-2011.)

Assertion

Ref	Expression
sbeqalbi	⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 = 𝑥 → 𝑧 = 𝑦))

Distinct variable groups:   𝑦,𝑧   𝑥,𝑧

Proof of Theorem sbeqalbi

Step	Hyp	Ref	Expression
1		equtrr 1949	. . 3 ⊢ (𝑥 = 𝑦 → (𝑧 = 𝑥 → 𝑧 = 𝑦))
2	1	alrimiv 1855	. 2 ⊢ (𝑥 = 𝑦 → ∀𝑧(𝑧 = 𝑥 → 𝑧 = 𝑦))
3		sbeqal1 38598	. 2 ⊢ (∀𝑧(𝑧 = 𝑥 → 𝑧 = 𝑦) → 𝑥 = 𝑦)
4	2, 3	impbii 199	1 ⊢ (𝑥 = 𝑦 ↔ ∀𝑧(𝑧 = 𝑥 → 𝑧 = 𝑦))

Syntax hints:   → wi 4   ↔ wb 196  ∀wal 1481

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)