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Metamath Proof Explorer

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 ) )