Theorem eximp-surprise2 50620

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 50619, which shows what implication really expands to. See also empty-surprise 50617. (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 881	. . 3 ⊢ (¬ 𝜑 → (¬ 𝜑 ∨ 𝜓))
3	1, 2	eximii 1870	. 2 ⊢ ∃𝑥(¬ 𝜑 ∨ 𝜓)
4		eximp-surprise 50619	. 2 ⊢ (∃𝑥(𝜑 → 𝜓) ↔ ∃𝑥(¬ 𝜑 ∨ 𝜓))
5	3, 4	mpbir 234	1 ⊢ ∃𝑥(𝜑 → 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∨ wo 861  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-or 862  df-ex 1813

This theorem is used by: (None)