Theorem mor 2061

Description: Converse of mo23 2060 with an additional ∃𝑥𝜑 condition. (Contributed by Jim Kingdon, 25-Jun-2018.)

Hypothesis

Ref	Expression
mor.1	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
mor	⊢ (∃𝑥𝜑 → (∀𝑥∀𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem mor

Step	Hyp	Ref	Expression
1		mor.1	. . 3 ⊢ Ⅎ𝑦𝜑
2	1	sb8e 1850	. 2 ⊢ (∃𝑥𝜑 ↔ ∃𝑦[𝑦 / 𝑥]𝜑)
3		impexp 261	. . . . 5 ⊢ (((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦) ↔ (𝜑 → ([𝑦 / 𝑥]𝜑 → 𝑥 = 𝑦)))
4		bi2.04 247	. . . . 5 ⊢ ((𝜑 → ([𝑦 / 𝑥]𝜑 → 𝑥 = 𝑦)) ↔ ([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)))
5	3, 4	bitri 183	. . . 4 ⊢ (((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦) ↔ ([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)))
6	5	2albii 1464	. . 3 ⊢ (∀𝑥∀𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦) ↔ ∀𝑥∀𝑦([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)))
7		nfs1v 1932	. . . . . 6 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑
8	7	nfri 1512	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
9	8	eximi 1593	. . . 4 ⊢ (∃𝑦[𝑦 / 𝑥]𝜑 → ∃𝑦∀𝑥[𝑦 / 𝑥]𝜑)
10		alim 1450	. . . . . . 7 ⊢ (∀𝑥([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)) → (∀𝑥[𝑦 / 𝑥]𝜑 → ∀𝑥(𝜑 → 𝑥 = 𝑦)))
11	10	alimi 1448	. . . . . 6 ⊢ (∀𝑦∀𝑥([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)) → ∀𝑦(∀𝑥[𝑦 / 𝑥]𝜑 → ∀𝑥(𝜑 → 𝑥 = 𝑦)))
12	11	a7s 1447	. . . . 5 ⊢ (∀𝑥∀𝑦([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)) → ∀𝑦(∀𝑥[𝑦 / 𝑥]𝜑 → ∀𝑥(𝜑 → 𝑥 = 𝑦)))
13		exim 1592	. . . . 5 ⊢ (∀𝑦(∀𝑥[𝑦 / 𝑥]𝜑 → ∀𝑥(𝜑 → 𝑥 = 𝑦)) → (∃𝑦∀𝑥[𝑦 / 𝑥]𝜑 → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
14	12, 13	syl 14	. . . 4 ⊢ (∀𝑥∀𝑦([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)) → (∃𝑦∀𝑥[𝑦 / 𝑥]𝜑 → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
15	9, 14	syl5com 29	. . 3 ⊢ (∃𝑦[𝑦 / 𝑥]𝜑 → (∀𝑥∀𝑦([𝑦 / 𝑥]𝜑 → (𝜑 → 𝑥 = 𝑦)) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
16	6, 15	syl5bi 151	. 2 ⊢ (∃𝑦[𝑦 / 𝑥]𝜑 → (∀𝑥∀𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
17	2, 16	sylbi 120	1 ⊢ (∃𝑥𝜑 → (∀𝑥∀𝑦((𝜑 ∧ [𝑦 / 𝑥]𝜑) → 𝑥 = 𝑦) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))

Syntax hints:   → wi 4   ∧ wa 103  ∀wal 1346  Ⅎwnf 1453  ∃wex 1485  [wsb 1755

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756

This theorem is referenced by:  modc  2062