Description: Shorter proof of eqeq1d based on more axioms. (Contributed by NM, 27-Dec-1993) (Revised by Wolf Lammen, 19-Nov-2019) (Proof modification is discouraged.) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Hypothesis | eqeq1d.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
| Assertion | eqeq1dALT | ⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | eqeq1d.1 | ⊢ ( 𝜑 → 𝐴 = 𝐵 ) | |
| 2 | dfcleq | ⊢ ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) | |
| 3 | 1 2 | sylib | ⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) |
| 4 | 3 | 19.21bi | ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) |
| 5 | 4 | bibi1d | ⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) ) |
| 6 | 5 | albidv | ⊢ ( 𝜑 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) ) |
| 7 | dfcleq | ⊢ ( 𝐴 = 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐶 ) ) | |
| 8 | dfcleq | ⊢ ( 𝐵 = 𝐶 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 ↔ 𝑥 ∈ 𝐶 ) ) | |
| 9 | 6 7 8 | 3bitr4g | ⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 ) ) |