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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