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Metamath Proof Explorer

Theorem sylan9eq

Description: An equality transitivity deduction. (Contributed by NM, 8-May-1994) (Proof shortened by Andrew Salmon, 25-May-2011)

Ref Expression
Hypotheses sylan9eq.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
sylan9eq.2 ⊢ ( 𝜓 → 𝐵 = 𝐶 )
Assertion sylan9eq ( ( 𝜑 ∧ 𝜓 ) → 𝐴 = 𝐶 )

Proof

Step Hyp Ref Expression
1 sylan9eq.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 sylan9eq.2 ⊢ ( 𝜓 → 𝐵 = 𝐶 )
3 eqtr ⊢ ( ( 𝐴 = 𝐵 ∧ 𝐵 = 𝐶 ) → 𝐴 = 𝐶 )
4 1 2 3 syl2an ⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝐴 = 𝐶 )