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Metamath Proof Explorer

Theorem sbccomlemOLD

Description: Obsolete version of sbccomlem as of 20-Aug-2025. (Contributed by NM, 14-Nov-2005) (Revised by Mario Carneiro, 18-Oct-2016) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion sbccomlemOLD ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ 𝐵 / 𝑦 ] [ 𝐴 / 𝑥 ] 𝜑 )

Proof

Step Hyp Ref Expression
1 excom ⊢ ( ∃ 𝑥 ∃ 𝑦 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ∃ 𝑦 ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
2 exdistr ⊢ ( ∃ 𝑥 ∃ 𝑦 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
3 an12 ⊢ ( ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ( 𝑦 = 𝐵 ∧ ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
4 3 exbii ⊢ ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ∃ 𝑥 ( 𝑦 = 𝐵 ∧ ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
5 19.42v ⊢ ( ∃ 𝑥 ( 𝑦 = 𝐵 ∧ ( 𝑥 = 𝐴 ∧ 𝜑 ) ) ↔ ( 𝑦 = 𝐵 ∧ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
6 4 5 bitri ⊢ ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ( 𝑦 = 𝐵 ∧ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
7 6 exbii ⊢ ( ∃ 𝑦 ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
8 1 2 7 3bitr3i ⊢ ( ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) ↔ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
9 sbc5 ⊢ ( [ 𝐴 / 𝑥 ] ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) ) )
10 sbc5 ⊢ ( [ 𝐵 / 𝑦 ] ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ↔ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) ) )
11 8 9 10 3bitr4i ⊢ ( [ 𝐴 / 𝑥 ] ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) ↔ [ 𝐵 / 𝑦 ] ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) )
12 sbc5 ⊢ ( [ 𝐵 / 𝑦 ] 𝜑 ↔ ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) )
13 12 sbcbii ⊢ ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] ∃ 𝑦 ( 𝑦 = 𝐵 ∧ 𝜑 ) )
14 sbc5 ⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) )
15 14 sbcbii ⊢ ( [ 𝐵 / 𝑦 ] [ 𝐴 / 𝑥 ] 𝜑 ↔ [ 𝐵 / 𝑦 ] ∃ 𝑥 ( 𝑥 = 𝐴 ∧ 𝜑 ) )
16 11 13 15 3bitr4i ⊢ ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝜑 ↔ [ 𝐵 / 𝑦 ] [ 𝐴 / 𝑥 ] 𝜑 )