Theorem eximp-surprise2 50447

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 50446, which shows what implication really expands to. See also empty-surprise 50444. (Contributed by David A. Wheeler, 18-Oct-2018.)

Hypothesis

Ref	Expression
eximp-surprise2.1	⊢ ∃𝑥 ¬ 𝜑

Assertion

Ref	Expression
eximp-surprise2	⊢ ∃𝑥(𝜑 → 𝜓)

Proof of Theorem eximp-surprise2

Step	Hyp	Ref	Expression
1		eximp-surprise2.1	. . 3 ⊢ ∃𝑥 ¬ 𝜑
2		orc 880	. . 3 ⊢ (¬ 𝜑 → (¬ 𝜑 ∨ 𝜓))
3	1, 2	eximii 1864	. 2 ⊢ ∃𝑥(¬ 𝜑 ∨ 𝜓)
4		eximp-surprise 50446	. 2 ⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓))
5	3, 4	mpbir 234	1 ⊢ ∃𝑥(𝜑 → 𝜓)

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 860  ∃wex 1806

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836

This theorem depends on definitions:  df-bi 210  df-or 861  df-ex 1807

This theorem is referenced by: (None)