Users' Mathboxes Mathbox for David A. Wheeler < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  alimp-surprise Structured version   Visualization version   GIF version

Theorem alimp-surprise 43632
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., 𝑥𝑀 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(𝜑𝜓). 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 (𝜑𝜓) is equivalent to ¬ 𝜑𝜓 (as proven in imor 842). 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 43633) or using the allsome quantifier (df-alsi 43640) .

For other "surprises" for new users of classical logic, see empty-surprise 43634 and eximp-surprise 43636. (Contributed by David A. Wheeler, 17-Oct-2018.)

Ref Expression
alimp-surprise.1 ¬ ∃𝑥𝜑
Ref Expression
alimp-surprise (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))

Proof of Theorem alimp-surprise
StepHypRef Expression
1 imor 842 . . . 4 ((𝜑𝜓) ↔ (¬ 𝜑𝜓))
21albii 1863 . . 3 (∀𝑥(𝜑𝜓) ↔ ∀𝑥𝜑𝜓))
3 alimp-surprise.1 . . . . 5 ¬ ∃𝑥𝜑
43nexr 2176 . . . 4 ¬ 𝜑
54orci 854 . . 3 𝜑𝜓)
62, 5mpgbir 1843 . 2 𝑥(𝜑𝜓)
7 imor 842 . . . 4 ((𝜑 → ¬ 𝜓) ↔ (¬ 𝜑 ∨ ¬ 𝜓))
87albii 1863 . . 3 (∀𝑥(𝜑 → ¬ 𝜓) ↔ ∀𝑥𝜑 ∨ ¬ 𝜓))
94orci 854 . . 3 𝜑 ∨ ¬ 𝜓)
108, 9mpgbir 1843 . 2 𝑥(𝜑 → ¬ 𝜓)
116, 10pm3.2i 464 1 (∀𝑥(𝜑𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 386  wo 836  wal 1599  wex 1823
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-12 2163
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-ex 1824
This theorem is referenced by:  empty-surprise  43634
  Copyright terms: Public domain W3C validator