Theorem equequ2 1599

Description: An equivalence law for equality. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
equequ2	⊢ (𝑥 = 𝑦 → (𝑧 = 𝑥 ↔ 𝑧 = 𝑦))

Proof of Theorem equequ2

Step	Hyp	Ref	Expression
1		equtrr 1596	. 2 ⊢ (𝑥 = 𝑦 → (𝑧 = 𝑥 → 𝑧 = 𝑦))
2		equtrr 1596	. . 3 ⊢ (𝑦 = 𝑥 → (𝑧 = 𝑦 → 𝑧 = 𝑥))
3	2	equcoms 1594	. 2 ⊢ (𝑥 = 𝑦 → (𝑧 = 𝑦 → 𝑧 = 𝑥))
4	1, 3	impbid 120	1 ⊢ (𝑥 = 𝑦 → (𝑧 = 𝑥 ↔ 𝑧 = 𝑦))

Syntax hints:   → wi 4   ↔ wb 98

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-gen 1338  ax-ie2 1383  ax-8 1395  ax-17 1419  ax-i9 1423

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  ax11v2  1701  ax11v  1708  ax11ev  1709  equs5or  1711  eujust  1902  euf  1905  mo23  1941  iotaval  4878  dffun4f  4918  dff13f  5409