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Metamath Proof Explorer

Theorem cbveuw

Description: Version of cbveu with a disjoint variable condition, which does not require ax-10 , ax-13 . (Contributed by NM, 25-Nov-1994) (Revised by GG, 23-May-2024)

Ref Expression
Hypotheses cbveuw.1 ⊢ Ⅎ 𝑦 𝜑
cbveuw.2 ⊢ Ⅎ 𝑥 𝜓
cbveuw.3 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion cbveuw ( ∃! 𝑥 𝜑 ↔ ∃! 𝑦 𝜓 )

Proof

Step Hyp Ref Expression
1 cbveuw.1 ⊢ Ⅎ 𝑦 𝜑
2 cbveuw.2 ⊢ Ⅎ 𝑥 𝜓
3 cbveuw.3 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
4 1 2 3 cbvexv1 ⊢ ( ∃ 𝑥 𝜑 ↔ ∃ 𝑦 𝜓 )
5 1 2 3 cbvmow ⊢ ( ∃* 𝑥 𝜑 ↔ ∃* 𝑦 𝜓 )
6 4 5 anbi12i ⊢ ( ( ∃ 𝑥 𝜑 ∧ ∃* 𝑥 𝜑 ) ↔ ( ∃ 𝑦 𝜓 ∧ ∃* 𝑦 𝜓 ) )
7 df-eu ⊢ ( ∃! 𝑥 𝜑 ↔ ( ∃ 𝑥 𝜑 ∧ ∃* 𝑥 𝜑 ) )
8 df-eu ⊢ ( ∃! 𝑦 𝜓 ↔ ( ∃ 𝑦 𝜓 ∧ ∃* 𝑦 𝜓 ) )
9 6 7 8 3bitr4i ⊢ ( ∃! 𝑥 𝜑 ↔ ∃! 𝑦 𝜓 )