Theorem wl-sbcom3 32551

Description: Substituting 𝑦 for 𝑥 and then 𝑧 for 𝑦 is equivalent to substituting 𝑧 for both 𝑥 and 𝑦. Copy of ~? sbcom3OLD with a shortened proof.

Keep this theorem for a while here because an external reference to it exists.

(Contributed by Giovanni Mascellani, 8-Apr-2018.) (Proof shortened by Wolf Lammen, 15-Sep-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
wl-sbcom3	⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)

Proof of Theorem wl-sbcom3

Step	Hyp	Ref	Expression
1		nfa1 2015	. . . 4 ⊢ Ⅎ𝑦∀𝑦 𝑦 = 𝑧
2		sbequ 2364	. . . . 5 ⊢ (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
3	2	sps 2043	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
4	1, 3	sbbid 2391	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
5	2	pm5.74i 259	. . . . . 6 ⊢ ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
6	5	albii 1737	. . . . 5 ⊢ (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
7	6	a1i 11	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
8		sb4b 2346	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
9		sb4b 2346	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
10	7, 8, 9	3bitr4d 299	. . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
11	4, 10	pm2.61i 175	. 2 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)
12		sbcom 2406	. 2 ⊢ ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)
13	11, 12	bitri 263	1 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195  ∀wal 1473  [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-11 2021  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)