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Theorem sbcom3 2528
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 2382. For a version requiring a disjoint variable, but fewer axioms, see sbcom3vv 2104. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 2159. (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
StepHypRef Expression
1 nfa1 2153 . . 3 𝑦𝑦 𝑦 = 𝑧
2 drsb2 2266 . . 3 (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
31, 2sbbid 2245 . 2 (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
4 sb4b 2491 . . . 4 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
5 sbequ 2089 . . . . . 6 (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
65pm5.74i 274 . . . . 5 ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
76albii 1821 . . . 4 (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
84, 7syl6bb 290 . . 3 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
9 sb4b 2491 . . 3 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
108, 9bitr4d 285 . 2 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
113, 10pm2.61i 185 1 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wal 1536  [wsb 2069
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2143  ax-12 2176  ax-13 2382
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-nf 1786  df-sb 2070
This theorem is referenced by:  sbco  2529  sbidm  2532  sbcom  2536
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