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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 42339, which shows what implication really expands to. See also empty-surprise 42337. (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 399 | . . 3 ⊢ (¬ 𝜑 → (¬ 𝜑 ∨ 𝜓)) | |
3 | 1, 2 | eximii 1754 | . 2 ⊢ ∃𝑥(¬ 𝜑 ∨ 𝜓) |
4 | eximp-surprise 42339 | . 2 ⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓)) | |
5 | 3, 4 | mpbir 220 | 1 ⊢ ∃𝑥(𝜑 → 𝜓) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 → wi 4 ∨ wo 382 ∃wex 1695 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 |
This theorem depends on definitions: df-bi 196 df-or 384 df-ex 1696 |
This theorem is referenced by: (None) |
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