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Metamath Proof Explorer

Theorem equvel

Description: A variable elimination law for equality with no distinct variable requirements. Compare equvini . Usage of this theorem is discouraged because it depends on ax-13 . Use equvelv when possible. (Contributed by NM, 1-Mar-2013) (Proof shortened by Mario Carneiro, 17-Oct-2016) (Proof shortened by Wolf Lammen, 15-Jun-2019) (New usage is discouraged.)

Ref Expression
Assertion equvel ( ∀ 𝑧 ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → 𝑥 = 𝑦 )

Proof

Step Hyp Ref Expression
1 albi ⊢ ( ∀ 𝑧 ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → ( ∀ 𝑧 𝑧 = 𝑥 ↔ ∀ 𝑧 𝑧 = 𝑦 ) )
2 ax6e ⊢ ∃ 𝑧 𝑧 = 𝑦
3 biimpr ⊢ ( ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → ( 𝑧 = 𝑦 → 𝑧 = 𝑥 ) )
4 ax7 ⊢ ( 𝑧 = 𝑥 → ( 𝑧 = 𝑦 → 𝑥 = 𝑦 ) )
5 3 4 syli ⊢ ( ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → ( 𝑧 = 𝑦 → 𝑥 = 𝑦 ) )
6 5 com12 ⊢ ( 𝑧 = 𝑦 → ( ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → 𝑥 = 𝑦 ) )
7 2 6 eximii ⊢ ∃ 𝑧 ( ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → 𝑥 = 𝑦 )
8 7 19.35i ⊢ ( ∀ 𝑧 ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → ∃ 𝑧 𝑥 = 𝑦 )
9 4 spsd ⊢ ( 𝑧 = 𝑥 → ( ∀ 𝑧 𝑧 = 𝑦 → 𝑥 = 𝑦 ) )
10 9 sps ⊢ ( ∀ 𝑧 𝑧 = 𝑥 → ( ∀ 𝑧 𝑧 = 𝑦 → 𝑥 = 𝑦 ) )
11 10 a1dd ⊢ ( ∀ 𝑧 𝑧 = 𝑥 → ( ∀ 𝑧 𝑧 = 𝑦 → ( ∃ 𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦 ) ) )
12 nfeqf ⊢ ( ( ¬ ∀ 𝑧 𝑧 = 𝑥 ∧ ¬ ∀ 𝑧 𝑧 = 𝑦 ) → Ⅎ 𝑧 𝑥 = 𝑦 )
13 12 19.9d ⊢ ( ( ¬ ∀ 𝑧 𝑧 = 𝑥 ∧ ¬ ∀ 𝑧 𝑧 = 𝑦 ) → ( ∃ 𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦 ) )
14 13 ex ⊢ ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( ¬ ∀ 𝑧 𝑧 = 𝑦 → ( ∃ 𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦 ) ) )
15 11 14 bija ⊢ ( ( ∀ 𝑧 𝑧 = 𝑥 ↔ ∀ 𝑧 𝑧 = 𝑦 ) → ( ∃ 𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦 ) )
16 1 8 15 sylc ⊢ ( ∀ 𝑧 ( 𝑧 = 𝑥 ↔ 𝑧 = 𝑦 ) → 𝑥 = 𝑦 )