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Metamath Proof Explorer

Theorem 2sb6rf

Description: Reversed double substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 3-Feb-2005) (Revised by Mario Carneiro, 6-Oct-2016) Remove variable constraints. (Revised by Wolf Lammen, 28-Sep-2018) (Proof shortened by Wolf Lammen, 13-Apr-2023) (New usage is discouraged.)

Ref Expression
Hypotheses 2sb5rf.1 ⊢ Ⅎ 𝑧 𝜑
2sb5rf.2 ⊢ Ⅎ 𝑤 𝜑
Assertion 2sb6rf ( 𝜑 ↔ ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 2sb5rf.1 ⊢ Ⅎ 𝑧 𝜑
2 2sb5rf.2 ⊢ Ⅎ 𝑤 𝜑
3 1 19.23 ⊢ ( ∀ 𝑧 ( ∃ 𝑤 ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) ↔ ( ∃ 𝑧 ∃ 𝑤 ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
4 2 19.23 ⊢ ( ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) ↔ ( ∃ 𝑤 ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
5 4 albii ⊢ ( ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) ↔ ∀ 𝑧 ( ∃ 𝑤 ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
6 2ax6e ⊢ ∃ 𝑧 ∃ 𝑤 ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 )
7 6 a1bi ⊢ ( 𝜑 ↔ ( ∃ 𝑧 ∃ 𝑤 ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
8 3 5 7 3bitr4ri ⊢ ( 𝜑 ↔ ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
9 sbequ12r ⊢ ( 𝑧 = 𝑥 → ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ [ 𝑤 / 𝑦 ] 𝜑 ) )
10 sbequ12r ⊢ ( 𝑤 = 𝑦 → ( [ 𝑤 / 𝑦 ] 𝜑 ↔ 𝜑 ) )
11 9 10 sylan9bb ⊢ ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ 𝜑 ) )
12 11 pm5.74i ⊢ ( ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
13 12 2albii ⊢ ( ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → 𝜑 ) )
14 8 13 bitr4i ⊢ ( 𝜑 ↔ ∀ 𝑧 ∀ 𝑤 ( ( 𝑧 = 𝑥 ∧ 𝑤 = 𝑦 ) → [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ) )