Theorem bj-sbsb 37719

Description: Biconditional showing two possible (dual) definitions of substitution df-sb 2100 not using dummy variables. (Contributed by BJ, 19-Mar-2021.)

Assertion

Ref	Expression
bj-sbsb	⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))

Proof of Theorem bj-sbsb

Step	Hyp	Ref	Expression
1		simpl 488	. . . 4 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → (𝑥 = 𝑦 → 𝜑))
2		pm2.27 43	. . . . . 6 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → 𝜑))
3	2	anc2li 565	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 ∧ 𝜑)))
4	3	sps 2222	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 ∧ 𝜑)))
5		olc 882	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))
6	1, 4, 5	syl56 37	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑))))
7		simpr 490	. . . 4 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8		equs5 2490	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
9	8	biimpd 232	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
10		orc 881	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))
11	7, 9, 10	syl56 37	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑))))
12	6, 11	pm2.61i 184	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))
13		sp 2220	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜑))
14		pm3.4 822	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → (𝑥 = 𝑦 → 𝜑))
15	13, 14	jaoi 871	. . 3 ⊢ ((∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)) → (𝑥 = 𝑦 → 𝜑))
16		equs4 2446	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
17		19.8a 2218	. . . 4 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
18	16, 17	jaoi 871	. . 3 ⊢ ((∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
19	15, 18	jca 521	. 2 ⊢ ((∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)) → ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
20	12, 19	impbii 212	1 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ (∀𝑥(𝑥 = 𝑦 → 𝜑) ∨ (𝑥 = 𝑦 ∧ 𝜑)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2178  ax-12 2213  ax-13 2402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817

This theorem is used by:  bj-dfsb2  37720