Theorem drsb1 2524

Description: Formula-building lemma for use with the Distinctor Reduction Theorem. Part of Theorem 9.4 of [Megill] p. 448 (p. 16 of preprint). Usage of this theorem is discouraged because it depends on ax-13 2401. (Contributed by NM, 2-Jun-1993.) (New usage is discouraged.)

Assertion

Ref	Expression
drsb1	⊢ (∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜑))

Proof of Theorem drsb1

Step	Hyp	Ref	Expression
1		equequ1 2057	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑦 = 𝑧))
2	1	sps 2221	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑦 = 𝑧))
3	2	imbi1d 344	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑧 → 𝜑) ↔ (𝑦 = 𝑧 → 𝜑)))
4	2	anbi1d 643	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑧 ∧ 𝜑) ↔ (𝑦 = 𝑧 ∧ 𝜑)))
5	4	drex1 2470	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑧 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝑧 ∧ 𝜑)))
6	3, 5	anbi12d 644	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑧 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑧 ∧ 𝜑)) ↔ ((𝑦 = 𝑧 → 𝜑) ∧ ∃𝑦(𝑦 = 𝑧 ∧ 𝜑))))
7		dfsb1 2510	. 2 ⊢ ([𝑧 / 𝑥]𝜑 ↔ ((𝑥 = 𝑧 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑧 ∧ 𝜑)))
8		dfsb1 2510	. 2 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ((𝑦 = 𝑧 → 𝜑) ∧ ∃𝑦(𝑦 = 𝑧 ∧ 𝜑)))
9	6, 7, 8	3bitr4g 317	1 ⊢ (∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑦]𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-10 2177  ax-12 2212  ax-13 2401

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-nf 1816  df-sb 2099

This theorem is used by:  sb2ae  2525  sbco3  2542  iotaeq  6509