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Metamath Proof Explorer

Theorem dfsb1

Description: Alternate definition of substitution. Remark 9.1 in Megill p. 447 (p. 15 of the preprint). This was the original definition before df-sb . Note that it does not require dummy variables in its definiens; this is done by having x free in the first conjunct and bound in the second. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by BJ, 9-Jul-2023) Revise df-sb . (Revised by Wolf Lammen, 29-Jul-2023) (New usage is discouraged.)

Ref Expression
Assertion dfsb1 ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 sbequ2 ⊢ ( 𝑥 = 𝑦 → ( [ 𝑦 / 𝑥 ] 𝜑 → 𝜑 ) )
2 1 com12 ⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ( 𝑥 = 𝑦 → 𝜑 ) )
3 sb1 ⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) )
4 2 3 jca ⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) )
5 id ⊢ ( 𝑥 = 𝑦 → 𝑥 = 𝑦 )
6 sbequ1 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜑 ) )
7 5 6 embantd ⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 = 𝑦 → 𝜑 ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
8 7 sps ⊢ ( ∀ 𝑥 𝑥 = 𝑦 → ( ( 𝑥 = 𝑦 → 𝜑 ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
9 8 adantrd ⊢ ( ∀ 𝑥 𝑥 = 𝑦 → ( ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
10 sb3 ⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
11 10 adantld ⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
12 9 11 pm2.61i ⊢ ( ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) → [ 𝑦 / 𝑥 ] 𝜑 )
13 4 12 impbii ⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) )