| Mathbox for Wolf Lammen |
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| Mirrors > Home > MPE Home > Th. List > Mathboxes > wl-df3maxtru1 | Structured version Visualization version GIF version | ||
| Description: Assuming
"(n+1)-maxtru1" ↔ ¬
"(n+1)-mintru-2", we can deduce from
the recursion formula given in wl-df-3mintru2 38514, that a similiar one
"(n+1)-maxtru1" ↔ if-(𝜑,-. "n-mintru-1" , "n-maxtru1" ) is valid for expressing 'at most one input is true'. This can also be rephrased as a mutual exclusivity of propositional expressions (no two of a sequence of inputs can simultaneously be true). Of course, this suggests that all inputs depend on variables 𝜂, 𝜁... Whatever wellformed expression we plugin for these variables, it will render at most one of the inputs true. The here introduced mutual exclusivity is possibly useful for case studies, where we want the cases be sort of 'disjoint'. One can further imagine that a complete case scenario demands that the 'at most' is sharpened to 'exactly one'. This does not impose any difficulty here, as one of the inputs will then be the negation of all others be or'ed. As one input is determined, 'at most one' is sufficient to describe the general form here. Since cadd is an alias for 'at least 2 out of three are true', its negation is under focus here. (Contributed by Wolf Lammen, 23-Jun-2024.) |
| Ref | Expression |
|---|---|
| wl-df3maxtru1 | ⊢ (¬ cadd(𝜑, 𝜓, 𝜒) ↔ if-(𝜑, (𝜓 ⊽ 𝜒), (𝜓 ⊼ 𝜒))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | cadnot 1647 | . 2 ⊢ (¬ cadd(𝜑, 𝜓, 𝜒) ↔ cadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒)) | |
| 2 | wl-df-3mintru2 38514 | . 2 ⊢ (cadd(¬ 𝜑, ¬ 𝜓, ¬ 𝜒) ↔ if-(¬ 𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜓 ∧ ¬ 𝜒))) | |
| 3 | ifpn 1090 | . . 3 ⊢ (if-(𝜑, (𝜓 ⊽ 𝜒), (𝜓 ⊼ 𝜒)) ↔ if-(¬ 𝜑, (𝜓 ⊼ 𝜒), (𝜓 ⊽ 𝜒))) | |
| 4 | nanor 1524 | . . . . . 6 ⊢ ((𝜓 ⊼ 𝜒) ↔ (¬ 𝜓 ∨ ¬ 𝜒)) | |
| 5 | 4 | a1i 11 | . . . . 5 ⊢ (⊤ → ((𝜓 ⊼ 𝜒) ↔ (¬ 𝜓 ∨ ¬ 𝜒))) |
| 6 | df-nor 1558 | . . . . . . 7 ⊢ ((𝜓 ⊽ 𝜒) ↔ ¬ (𝜓 ∨ 𝜒)) | |
| 7 | ioran 999 | . . . . . . 7 ⊢ (¬ (𝜓 ∨ 𝜒) ↔ (¬ 𝜓 ∧ ¬ 𝜒)) | |
| 8 | 6, 7 | bitri 278 | . . . . . 6 ⊢ ((𝜓 ⊽ 𝜒) ↔ (¬ 𝜓 ∧ ¬ 𝜒)) |
| 9 | 8 | a1i 11 | . . . . 5 ⊢ (⊤ → ((𝜓 ⊽ 𝜒) ↔ (¬ 𝜓 ∧ ¬ 𝜒))) |
| 10 | 5, 9 | ifpbi23d 1096 | . . . 4 ⊢ (⊤ → (if-(¬ 𝜑, (𝜓 ⊼ 𝜒), (𝜓 ⊽ 𝜒)) ↔ if-(¬ 𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜓 ∧ ¬ 𝜒)))) |
| 11 | 10 | mptru 1576 | . . 3 ⊢ (if-(¬ 𝜑, (𝜓 ⊼ 𝜒), (𝜓 ⊽ 𝜒)) ↔ if-(¬ 𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜓 ∧ ¬ 𝜒))) |
| 12 | 3, 11 | bitr2i 279 | . 2 ⊢ (if-(¬ 𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜓 ∧ ¬ 𝜒)) ↔ if-(𝜑, (𝜓 ⊽ 𝜒), (𝜓 ⊼ 𝜒))) |
| 13 | 1, 2, 12 | 3bitri 300 | 1 ⊢ (¬ cadd(𝜑, 𝜓, 𝜒) ↔ if-(𝜑, (𝜓 ⊽ 𝜒), (𝜓 ⊼ 𝜒))) |
| Colors of variables: wff setvar class |
| This proof depends on syntax axioms: ¬ wn 3 ↔ wb 209 ∧ wa 401 ∨ wo 861 if-wif 1078 ⊼ wnan 1520 ⊽ wnor 1557 ⊤wtru 1570 caddwcad 1638 |
| This proof depends on axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 |
| This proof depends on definitions: df-bi 210 df-an 402 df-or 862 df-ifp 1079 df-3or 1104 df-3an 1105 df-nan 1521 df-xor 1541 df-nor 1558 df-tru 1572 df-cad 1639 |
| This theorem is used by: (None) |
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