Theorem eqeq1d

Description: Deduction from equality to equivalence of equalities. (Contributed by NM, 27-Dec-1993) Reduce dependencies on axioms. (Revised by Wolf Lammen, 5-Dec-2019)

	Ref	Expression
Hypothesis	eqeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	eqeq1d	⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		eqeq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		dfcleq	⊢ ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
3	2	biimpi	⊢ ( 𝐴 = 𝐵 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
4		bibi1	⊢ ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) )
5	4	alimi	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) → ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) )
6		albi	⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) )
7	1 3 5 6	4syl	⊢ ( 𝜑 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) )
8		dfcleq	⊢ ( 𝐴 = 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) )
9		dfcleq	⊢ ( 𝐵 = 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) )
10	7 8 9	3bitr4g	⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 ) )