Theorem sbcom3 2514

Description: Substituting 𝑦 for 𝑥 and then 𝑧 for 𝑦 is equivalent to substituting 𝑧 for both 𝑥 and 𝑦. Usage of this theorem is discouraged because it depends on ax-13 2380. For a version requiring a disjoint variable, but fewer axioms, see sbcom3vv 2097. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 2158. (Revised by Wolf Lammen, 16-Sep-2018.) (Proof shortened by Wolf Lammen, 16-Sep-2018.) (New usage is discouraged.)

Assertion

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

Proof of Theorem sbcom3

Step	Hyp	Ref	Expression
1		nfa1 2152	. . 3 ⊢ Ⅎ𝑦∀𝑦 𝑦 = 𝑧
2		drsb2 2267	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
3	1, 2	sbbid 2247	. 2 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
4		sb4b 2483	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
5		sbequ 2083	. . . . . 6 ⊢ (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
6	5	pm5.74i 271	. . . . 5 ⊢ ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
7	6	albii 1817	. . . 4 ⊢ (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
8	4, 7	bitrdi 287	. . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
9		sb4b 2483	. . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
10	8, 9	bitr4d 282	. 2 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
11	3, 10	pm2.61i 182	1 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206  ∀wal 1535  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-10 2141  ax-12 2178  ax-13 2380

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-ex 1778  df-nf 1782  df-sb 2065

This theorem is referenced by:  sbco  2515  sbidm  2518  sbcom  2522