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Metamath Proof Explorer

Theorem sbcom3

Description: Substituting y for x and then z for y is equivalent to substituting z for both x and y . Usage of this theorem is discouraged because it depends on ax-13 . For a version requiring a disjoint variable, but fewer axioms, see sbcom3vv . (Contributed by Giovanni Mascellani, 8-Apr-2018) Remove dependency on ax-11 . (Revised by Wolf Lammen, 16-Sep-2018) (Proof shortened by Wolf Lammen, 16-Sep-2018) (New usage is discouraged.)

Ref Expression
Assertion sbcom3 ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 )

Proof

Step Hyp Ref Expression
1 nfa1 ⊢ Ⅎ 𝑦 ∀ 𝑦 𝑦 = 𝑧
2 drsb2 ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑥 ] 𝜑 ) )
3 1 2 sbbid ⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 ) )
4 sb4b ⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑦 / 𝑥 ] 𝜑 ) ) )
5 sbequ ⊢ ( 𝑦 = 𝑧 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑥 ] 𝜑 ) )
6 5 pm5.74i ⊢ ( ( 𝑦 = 𝑧 → [ 𝑦 / 𝑥 ] 𝜑 ) ↔ ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) )
7 6 albii ⊢ ( ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑦 / 𝑥 ] 𝜑 ) ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) )
8 4 7 bitrdi ⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) ) )
9 sb4b ⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → [ 𝑧 / 𝑥 ] 𝜑 ) ) )
10 8 9 bitr4d ⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 ) )
11 3 10 pm2.61i ⊢ ( [ 𝑧 / 𝑦 ] [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] [ 𝑧 / 𝑥 ] 𝜑 )