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Metamath Proof Explorer

Theorem ceqsrex2v

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by NM, 29-Oct-2005)

Ref Expression
Hypotheses ceqsrex2v.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
ceqsrex2v.2 ⊢ ( 𝑦 = 𝐵 → ( 𝜓 ↔ 𝜒 ) )
Assertion ceqsrex2v ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( ∃ 𝑥 ∈ 𝐶 ∃ 𝑦 ∈ 𝐷 ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 ceqsrex2v.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 ceqsrex2v.2 ⊢ ( 𝑦 = 𝐵 → ( 𝜓 ↔ 𝜒 ) )
3 anass ⊢ ( ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
4 3 rexbii ⊢ ( ∃ 𝑦 ∈ 𝐷 ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ ∃ 𝑦 ∈ 𝐷 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
5 r19.42v ⊢ ( ∃ 𝑦 ∈ 𝐷 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
6 4 5 bitri ⊢ ( ∃ 𝑦 ∈ 𝐷 ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
7 6 rexbii ⊢ ( ∃ 𝑥 ∈ 𝐶 ∃ 𝑦 ∈ 𝐷 ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ ∃ 𝑥 ∈ 𝐶 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
8 1 anbi2d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑦 = 𝐵 ∧ 𝜑 ) ↔ ( 𝑦 = 𝐵 ∧ 𝜓 ) ) )
9 8 rexbidv ⊢ ( 𝑥 = 𝐴 → ( ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜑 ) ↔ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜓 ) ) )
10 9 ceqsrexv ⊢ ( 𝐴 ∈ 𝐶 → ( ∃ 𝑥 ∈ 𝐶 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜓 ) ) )
11 7 10 bitrid ⊢ ( 𝐴 ∈ 𝐶 → ( ∃ 𝑥 ∈ 𝐶 ∃ 𝑦 ∈ 𝐷 ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜓 ) ) )
12 2 ceqsrexv ⊢ ( 𝐵 ∈ 𝐷 → ( ∃ 𝑦 ∈ 𝐷 ( 𝑦 = 𝐵 ∧ 𝜓 ) ↔ 𝜒 ) )
13 11 12 sylan9bb ⊢ ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( ∃ 𝑥 ∈ 𝐶 ∃ 𝑦 ∈ 𝐷 ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) ∧ 𝜑 ) ↔ 𝜒 ) )