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Theorem sbcom3 2572
Description: Substituting 𝑦 for 𝑥 and then 𝑧 for 𝑦 is equivalent to substituting 𝑧 for both 𝑥 and 𝑦. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 2202. (Revised by Wolf Lammen, 16-Sep-2018.) (Proof shortened by Wolf Lammen, 16-Sep-2018.)
Assertion
Ref Expression
sbcom3 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)

Proof of Theorem sbcom3
StepHypRef Expression
1 nfa1 2196 . . 3 𝑦𝑦 𝑦 = 𝑧
2 drsb2 2539 . . 3 (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
31, 2sbbid 2564 . 2 (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
4 sb4b 2519 . . . 4 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
5 sbequ 2537 . . . . . 6 (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
65pm5.74i 262 . . . . 5 ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
76albii 1904 . . . 4 (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
84, 7syl6bb 278 . . 3 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
9 sb4b 2519 . . 3 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
108, 9bitr4d 273 . 2 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
113, 10pm2.61i 176 1 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 197  wal 1635  [wsb 2061
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2069  ax-7 2105  ax-10 2186  ax-12 2215  ax-13 2422
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-ex 1860  df-nf 1864  df-sb 2062
This theorem is referenced by:  sbco  2573  sbidm  2575  sbcom  2579  equsb3  2594  elsb3  2598  elsb4  2601  wl-equsb3  33654
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