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Metamath Proof Explorer

Theorem 3eqtr4d

Description: A deduction from three chained equalities. (Contributed by NM, 4-Aug-1995) (Proof shortened by Andrew Salmon, 25-May-2011)

Ref Expression
Hypotheses 3eqtr4d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
3eqtr4d.2 ⊢ ( 𝜑 → 𝐶 = 𝐴 )
3eqtr4d.3 ⊢ ( 𝜑 → 𝐷 = 𝐵 )
Assertion 3eqtr4d ( 𝜑 → 𝐶 = 𝐷 )

Proof

Step Hyp Ref Expression
1 3eqtr4d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 3eqtr4d.2 ⊢ ( 𝜑 → 𝐶 = 𝐴 )
3 3eqtr4d.3 ⊢ ( 𝜑 → 𝐷 = 𝐵 )
4 3 1 eqtr4d ⊢ ( 𝜑 → 𝐷 = 𝐴 )
5 2 4 eqtr4d ⊢ ( 𝜑 → 𝐶 = 𝐷 )