Theorem wl-sbcom3 33372

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 2028	. . . 4 ⊢ Ⅎ𝑦∀𝑦 𝑦 = 𝑧
2		sbequ 2376	. . . . 5 ⊢ (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
3	2	sps 2055	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
4	1, 3	sbbid 2403	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
5	2	pm5.74i 260	. . . . . 6 ⊢ ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
6	5	albii 1747	. . . . 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 2358	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
9		sb4b 2358	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
10	7, 8, 9	3bitr4d 300	. . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
11	4, 10	pm2.61i 176	. 2 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)
12		sbcom 2418	. 2 ⊢ ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)
13	11, 12	bitri 264	1 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196  ∀wal 1481  [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)