Theorem sbcom3 2548

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 2390. For a version requiring a disjoint variable, but fewer axioms, see sbcom3vv 2106. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 2161. (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 2155	. . 3 ⊢ Ⅎ𝑦∀𝑦 𝑦 = 𝑧
2		drsb2 2267	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
3	1, 2	sbbid 2246	. 2 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
4		sb4b 2499	. . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
5		sbequ 2090	. . . . . 6 ⊢ (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
6	5	pm5.74i 273	. . . . 5 ⊢ ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
7	6	albii 1820	. . . 4 ⊢ (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
8	4, 7	syl6bb 289	. . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
9		sb4b 2499	. . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
10	8, 9	bitr4d 284	. 2 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
11	3, 10	pm2.61i 184	1 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208  ∀wal 1535  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2145  ax-12 2177  ax-13 2390

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-nf 1785  df-sb 2070

This theorem is referenced by:  sbco  2549  sbidm  2552  sbcom  2556