Theorem sb5ALT 37752

Description: Equivalence for substitution. Alternate proof of sb5 2418. This proof is sb5ALTVD 38171 automatically translated and minimized. (Contributed by Alan Sare, 21-Apr-2013.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sb5ALT	⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem sb5ALT

Step	Hyp	Ref	Expression
1		equsb1 2356	. . . 4 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
2		sban 2387	. . . . 5 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
3	2	simplbi2com 655	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)))
4	1, 3	mpi 20	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑))
5		spsbe 1871	. . 3 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
6	4, 5	syl 17	. 2 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7		hbs1 2424	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
8		simpr 476	. . . . 5 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜑)
9	8	a1i 11	. . . 4 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ((𝑥 = 𝑦 ∧ 𝜑) → 𝜑))
10		simpl 472	. . . . 5 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝑥 = 𝑦)
11	10	a1i 11	. . . 4 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ((𝑥 = 𝑦 ∧ 𝜑) → 𝑥 = 𝑦))
12		sbequ1 2096	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
13	12	com12 32	. . . 4 ⊢ (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
14	9, 11, 13	syl6c 68	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ((𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑))
15	7, 14	exlimexi 37751	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
16	6, 15	impbii 198	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∃wex 1695  [wsb 1867

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-12 2034  ax-13 2234

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868

This theorem is referenced by: (None)