Theorem alimp-surprise 50590

Description: Demonstrate that when using "for all" and material implication the consequent can be both always true and always false if there is no case where the antecedent is true.

Those inexperienced with formal notations of classical logic can be surprised with what "for all" and material implication do together when the implication's antecedent is never true. This can happen, for example, when the antecedent is set membership but the set is the empty set (e.g., 𝑥 ∈ 𝑀 and 𝑀 = ∅).

This is perhaps best explained using an example. The sentence "All Martians are green" would typically be represented formally using the expression ∀𝑥(𝜑 → 𝜓). In this expression 𝜑 is true iff 𝑥 is a Martian and 𝜓 is true iff 𝑥 is green. Similarly, "All Martians are not green" would typically be represented as ∀𝑥(𝜑 → ¬ 𝜓). However, if there are no Martians (¬ ∃𝑥𝜑), then both of those expressions are true. That is surprising to the inexperienced, because the two expressions seem to be the opposite of each other. The reason this occurs is because in classical logic the implication (𝜑 → 𝜓) is equivalent to ¬ 𝜑 ∨ 𝜓 (as proven in imor 867). When 𝜑 is always false, ¬ 𝜑 is always true, and an or with true is always true.

Here are a few technical notes. In this notation, 𝜑 and 𝜓 are predicates that return a true or false value and may depend on 𝑥. We only say may because it actually doesn't matter for our proof. In Metamath this simply means that we do not require that 𝜑, 𝜓, and 𝑥 be distinct (so 𝑥 can be part of 𝜑 or 𝜓).

In natural language the term "implies" often presumes that the antecedent can occur in at one least circumstance and that there is some sort of causality. However, exactly what causality means is complex and situation-dependent. Modern logic typically uses material implication instead; this has a rigorous definition, but it is important for new users of formal notation to precisely understand it. There are ways to solve this, e.g., expressly stating that the antecedent exists (see alimp-no-surprise 50591) or using the allsome quantifier (df-als 50598) .

For other "surprises" for new users of classical logic, see empty-surprise 50592 and eximp-surprise 50594. (Contributed by David A. Wheeler, 17-Oct-2018.)

Hypothesis

Ref	Expression
alimp-surprise.1	⊢ ¬ ∃𝑥𝜑

Assertion

Ref	Expression
alimp-surprise	⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))

Proof of Theorem alimp-surprise

Step	Hyp	Ref	Expression
1		imor 867	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ (¬ 𝜑 ∨ 𝜓))
2	1	albii 1852	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) ↔ ∀𝑥(¬ 𝜑 ∨ 𝜓))
3		alimp-surprise.1	. . . . 5 ⊢ ¬ ∃𝑥𝜑
4	3	nexr 2231	. . . 4 ⊢ ¬ 𝜑
5	4	orci 879	. . 3 ⊢ (¬ 𝜑 ∨ 𝜓)
6	2, 5	mpgbir 1832	. 2 ⊢ ∀𝑥(𝜑 → 𝜓)
7		imor 867	. . . 4 ⊢ ((𝜑 → ¬ 𝜓) ↔ (¬ 𝜑 ∨ ¬ 𝜓))
8	7	albii 1852	. . 3 ⊢ (∀𝑥(𝜑 → ¬ 𝜓) ↔ ∀𝑥(¬ 𝜑 ∨ ¬ 𝜓))
9	4	orci 879	. . 3 ⊢ (¬ 𝜑 ∨ ¬ 𝜓)
10	8, 9	mpgbir 1832	. 2 ⊢ ∀𝑥(𝜑 → ¬ 𝜓)
11	6, 10	pm3.2i 476	1 ⊢ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861  ∀wal 1568  ∃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  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813

This theorem is used by:  empty-surprise  50592