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Mirrors > Home > MPE Home > Th. List > sbcom3 | Structured version Visualization version GIF version |
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 2375. For a version requiring a disjoint variable, but fewer axioms, see sbcom3vv 2095. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 2155. (Revised by Wolf Lammen, 16-Sep-2018.) (Proof shortened by Wolf Lammen, 16-Sep-2018.) (New usage is discouraged.) |
Ref | Expression |
---|---|
sbcom3 | ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | nfa1 2149 | . . 3 ⊢ Ⅎ𝑦∀𝑦 𝑦 = 𝑧 | |
2 | drsb2 2264 | . . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑)) | |
3 | 1, 2 | sbbid 2244 | . 2 ⊢ (∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)) |
4 | sb4b 2478 | . . . 4 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑))) | |
5 | sbequ 2081 | . . . . . 6 ⊢ (𝑦 = 𝑧 → ([𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑥]𝜑)) | |
6 | 5 | pm5.74i 271 | . . . . 5 ⊢ ((𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ (𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)) |
7 | 6 | albii 1816 | . . . 4 ⊢ (∀𝑦(𝑦 = 𝑧 → [𝑦 / 𝑥]𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑)) |
8 | 4, 7 | bitrdi 287 | . . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))) |
9 | sb4b 2478 | . . 3 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑧 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → [𝑧 / 𝑥]𝜑))) | |
10 | 8, 9 | bitr4d 282 | . 2 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑)) |
11 | 3, 10 | pm2.61i 182 | 1 ⊢ ([𝑧 / 𝑦][𝑦 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝑧 / 𝑥]𝜑) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ↔ wb 206 ∀wal 1535 [wsb 2062 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-10 2139 ax-12 2175 ax-13 2375 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-ex 1777 df-nf 1781 df-sb 2063 |
This theorem is referenced by: sbco 2510 sbidm 2513 sbcom 2517 |
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