Description: Equality implies equivalence with substitution. (Contributed by NM, 2-Mar-1995) (Proof shortened by BJ, 1-May-2019)
Ref | Expression | ||
---|---|---|---|
Assertion | ceqex | ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 19.8a | ⊢ ( ( 𝑥 = 𝐴 ∧ 𝜑 ) → ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) | |
2 | 1 | ex | ⊢ ( 𝑥 = 𝐴 → ( 𝜑 → ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) ) |
3 | eqvisset | ⊢ ( 𝑥 = 𝐴 → 𝐴 ∈ V ) | |
4 | alexeqg | ⊢ ( 𝐴 ∈ V → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) ) | |
5 | 3 4 | syl | ⊢ ( 𝑥 = 𝐴 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) ) |
6 | sp | ⊢ ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) → ( 𝑥 = 𝐴 → 𝜑 ) ) | |
7 | 6 | com12 | ⊢ ( 𝑥 = 𝐴 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝜑 ) → 𝜑 ) ) |
8 | 5 7 | sylbird | ⊢ ( 𝑥 = 𝐴 → ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) → 𝜑 ) ) |
9 | 2 8 | impbid | ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) ) |