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Metamath Proof Explorer

Theorem equsal

Description: An equivalence related to implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 . See equsalvw and equsalv for versions with disjoint variable conditions proved from fewer axioms. See also the dual form equsex . (Contributed by NM, 2-Jun-1993) (Proof shortened by Andrew Salmon, 12-Aug-2011) (Revised by Mario Carneiro, 3-Oct-2016) (Proof shortened by Wolf Lammen, 5-Feb-2018) (New usage is discouraged.)

Ref Expression
Hypotheses equsal.1 ⊢ Ⅎ 𝑥 𝜓
equsal.2 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion equsal ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ 𝜓 )

Proof

Step Hyp Ref Expression
1 equsal.1 ⊢ Ⅎ 𝑥 𝜓
2 equsal.2 ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
3 1 19.23 ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) ↔ ( ∃ 𝑥 𝑥 = 𝑦 → 𝜓 ) )
4 2 pm5.74i ⊢ ( ( 𝑥 = 𝑦 → 𝜑 ) ↔ ( 𝑥 = 𝑦 → 𝜓 ) )
5 4 albii ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) )
6 ax6e ⊢ ∃ 𝑥 𝑥 = 𝑦
7 6 a1bi ⊢ ( 𝜓 ↔ ( ∃ 𝑥 𝑥 = 𝑦 → 𝜓 ) )
8 3 5 7 3bitr4i ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ 𝜓 )