Theorem sbeqal1i 42017

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 42016	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝑥 = 𝑧) → 𝑦 = 𝑧)
2		sbeqal1i.1	. 2 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑧)
3	1, 2	mpg 1800	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 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171  ax-13 2372

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1783  df-nf 1787  df-sb 2068

This theorem is referenced by:  sbeqal2i  42018