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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