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Metamath Proof Explorer

Theorem 2sb6

Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005)

Ref Expression
Assertion 2sb6 ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )

Proof

Step Hyp Ref Expression
1 sb6 ⊢ ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) )
2 19.21v ⊢ ( ∀ 𝑦 ( 𝑥 = 𝑧 → ( 𝑦 = 𝑤 → 𝜑 ) ) ↔ ( 𝑥 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜑 ) ) )
3 impexp ⊢ ( ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) ↔ ( 𝑥 = 𝑧 → ( 𝑦 = 𝑤 → 𝜑 ) ) )
4 3 albii ⊢ ( ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) ↔ ∀ 𝑦 ( 𝑥 = 𝑧 → ( 𝑦 = 𝑤 → 𝜑 ) ) )
5 sb6 ⊢ ( [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜑 ) )
6 5 imbi2i ⊢ ( ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ( 𝑥 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜑 ) ) )
7 2 4 6 3bitr4ri ⊢ ( ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )
8 7 albii ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑧 → [ 𝑤 / 𝑦 ] 𝜑 ) ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )
9 1 8 bitri ⊢ ( [ 𝑧 / 𝑥 ] [ 𝑤 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( ( 𝑥 = 𝑧 ∧ 𝑦 = 𝑤 ) → 𝜑 ) )