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Theorem sbco3 2555
Description: A composition law for substitution. Usage of this theorem is discouraged because it depends on ax-13 2390. (Contributed by NM, 2-Jun-1993.) (Proof shortened by Wolf Lammen, 18-Sep-2018.) (New usage is discouraged.)
Assertion
Ref Expression
sbco3 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑥 / 𝑦]𝜑)

Proof of Theorem sbco3
StepHypRef Expression
1 drsb1 2535 . . 3 (∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑥][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑦 / 𝑥]𝜑))
2 nfae 2455 . . . 4 𝑥𝑥 𝑥 = 𝑦
3 sbequ12a 2256 . . . . 5 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ [𝑥 / 𝑦]𝜑))
43sps 2184 . . . 4 (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 ↔ [𝑥 / 𝑦]𝜑))
52, 4sbbid 2246 . . 3 (∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑥][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑥 / 𝑦]𝜑))
61, 5bitr3d 283 . 2 (∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑥 / 𝑦]𝜑))
7 sbco 2549 . . . 4 ([𝑥 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑥 / 𝑦]𝜑)
87sbbii 2081 . . 3 ([𝑧 / 𝑥][𝑥 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑥 / 𝑦]𝜑)
9 nfnae 2456 . . . 4 𝑦 ¬ ∀𝑥 𝑥 = 𝑦
10 nfnae 2456 . . . 4 𝑥 ¬ ∀𝑥 𝑥 = 𝑦
11 nfsb2 2522 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → Ⅎ𝑥[𝑦 / 𝑥]𝜑)
129, 10, 11sbco2d 2554 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑥][𝑥 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑦 / 𝑥]𝜑))
138, 12syl5rbbr 288 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑥 / 𝑦]𝜑))
146, 13pm2.61i 184 1 ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥][𝑥 / 𝑦]𝜑)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  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-11 2161  ax-12 2177  ax-13 2390
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070
This theorem is referenced by:  sbcom  2556
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