Theorem hbeu 1387

Description: Bound-variable hypothesis builder for "at most one." Note that x and y needn't be distinct (this makes the proof more difficult).

Hypothesis

Ref	Expression
hbeu.1	⊢ (φ → ∀xφ)

Assertion

Ref	Expression
hbeu	⊢ (∃!yφ → ∀x∃!yφ)

Proof of Theorem hbeu

Step	Hyp	Ref	Expression
1		ax-10o 1138	. . . . . 6 ⊢ (∀y y = x → (∀y∀y(φ ↔ y = z) → ∀x∀y(φ ↔ y = z)))
2	1	alequcoms 1141	. . . . 5 ⊢ (∀x x = y → (∀y∀y(φ ↔ y = z) → ∀x∀y(φ ↔ y = z)))
3		hba1 1001	. . . . 5 ⊢ (∀y(φ ↔ y = z) → ∀y∀y(φ ↔ y = z))
4	2, 3	syl5 21	. . . 4 ⊢ (∀x x = y → (∀y(φ ↔ y = z) → ∀x∀y(φ ↔ y = z)))
5		hbnae 1145	. . . . 5 ⊢ (¬ ∀x x = y → ∀y ¬ ∀x x = y)
6		hbnae 1145	. . . . . 6 ⊢ (¬ ∀x x = y → ∀x ¬ ∀x x = y)
7		hbeu.1	. . . . . . 7 ⊢ (φ → ∀xφ)
8	7	a1i 8	. . . . . 6 ⊢ (¬ ∀x x = y → (φ → ∀xφ))
9		dveeq1 1352	. . . . . 6 ⊢ (¬ ∀x x = y → (y = z → ∀x y = z))
10	6, 8, 9	hbbid 1110	. . . . 5 ⊢ (¬ ∀x x = y → ((φ ↔ y = z) → ∀x(φ ↔ y = z)))
11	5, 10	hbald 1111	. . . 4 ⊢ (¬ ∀x x = y → (∀y(φ ↔ y = z) → ∀x∀y(φ ↔ y = z)))
12	4, 11	pm2.61i 126	. . 3 ⊢ (∀y(φ ↔ y = z) → ∀x∀y(φ ↔ y = z))
13	12	hbex 1004	. 2 ⊢ (∃z∀y(φ ↔ y = z) → ∀x∃z∀y(φ ↔ y = z))
14		df-eu 1380	. 2 ⊢ (∃!yφ ↔ ∃z∀y(φ ↔ y = z))
15	14	albii 997	. 2 ⊢ (∀x∃!yφ ↔ ∀x∃z∀y(φ ↔ y = z))
16	13, 14, 15	3imtr4 219	1 ⊢ (∃!yφ → ∀x∃!yφ)

Syntax hints:  ¬ wn 2   → wi 3   ↔ wb 146  ∀wal 952  ∃wex 978  ∃!weu 1378

This theorem is referenced by:  hbmo 1405  2eu7 1453  2eu8 1454

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-eu 1380

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