Theorem sbeqal1i 38599

Description: Suppose you know 𝑥 = 𝑦 implies 𝑥 = 𝑧, assuming 𝑥 and 𝑧 are distinct. Then, 𝑦 = 𝑧. (Contributed by Andrew Salmon, 3-Jun-2011.)

Hypothesis

Ref	Expression
sbeqal1i.1	⊢ (𝑥 = 𝑦 → 𝑥 = 𝑧)

Assertion

Ref	Expression
sbeqal1i	⊢ 𝑦 = 𝑧

Distinct variable group:   𝑥,𝑧

Proof of Theorem sbeqal1i

Step	Hyp	Ref	Expression
1		sbeqal1 38598	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧) → 𝑦 = 𝑧)
2		sbeqal1i.1	. 2 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑧)
3	1, 2	mpg 1724	1 ⊢ 𝑦 = 𝑧

Syntax hints:   → wi 4

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:  sbeqal2i  38600