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Theorem wl-sbcom3 32351
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
StepHypRef Expression
1 nfa1 2014 . . . 4 𝑦𝑦 𝑦 = 𝑧
2 sbequ 2360 . . . . 5 (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
32sps 2041 . . . 4 (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑))
41, 3sbbid 2387 . . 3 (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
52pm5.74i 258 . . . . . 6 ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
65albii 1736 . . . . 5 (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))
76a1i 11 . . . 4 (¬ ∀𝑦 𝑦 = 𝑧 → (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
8 sb4b 2342 . . . 4 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑)))
9 sb4b 2342 . . . 4 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)))
107, 8, 93bitr4d 298 . . 3 (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑))
114, 10pm2.61i 174 . 2 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)
12 sbcom 2402 . 2 ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)
1311, 12bitri 262 1 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑧 / 𝑦]𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 194  wal 1472  [wsb 1866
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-11 2020  ax-12 2032  ax-13 2229
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867
This theorem is referenced by: (None)
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