Mathbox for David A. Wheeler |
< Previous
Next >
Nearby theorems |
||
Mirrors > Home > MPE Home > Th. List > Mathboxes > eximp-surprise2 | Structured version Visualization version GIF version |
Description: Show that "there
exists" with an implication is always true if there
exists a situation where the antecedent is false.
Those inexperienced with formal notations of classical logic may use expressions combining "there exists" with implication. This is usually a mistake, because that combination does not mean what an inexperienced person might think it means. For example, if there is some object that does not meet the precondition 𝜑, then the expression ∃𝑥(𝜑 → 𝜓) as a whole is always true, no matter what 𝜓 is (𝜓 could even be false, ⊥). New users of formal notation who use "there exists" with an implication should consider if they meant "and" instead of "implies". See eximp-surprise 41863, which shows what implication really expands to. See also empty-surprise 41861. (Contributed by David A. Wheeler, 18-Oct-2018.) |
Ref | Expression |
---|---|
eximp-surprise2.1 | ⊢ ∃𝑥 ¬ 𝜑 |
Ref | Expression |
---|---|
eximp-surprise2 | ⊢ ∃𝑥(𝜑 → 𝜓) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eximp-surprise2.1 | . . 3 ⊢ ∃𝑥 ¬ 𝜑 | |
2 | orc 400 | . . 3 ⊢ (¬ 𝜑 → (¬ 𝜑 ∨ 𝜓)) | |
3 | 1, 2 | eximii 1761 | . 2 ⊢ ∃𝑥(¬ 𝜑 ∨ 𝜓) |
4 | eximp-surprise 41863 | . 2 ⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓)) | |
5 | 3, 4 | mpbir 221 | 1 ⊢ ∃𝑥(𝜑 → 𝜓) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∨ wo 383 ∃wex 1701 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 |
This theorem depends on definitions: df-bi 197 df-or 385 df-ex 1702 |
This theorem is referenced by: (None) |
Copyright terms: Public domain | W3C validator |