Theorem alimp-no-surprise 46485

Description: There is no "surprise" in a for-all with implication if there exists a value where the antecedent is true. This is one way to prevent for-all with implication from allowing anything. For a contrast, see alimp-surprise 46484. The allsome quantifier also counters this problem, see df-alsi 46492. (Contributed by David A. Wheeler, 27-Oct-2018.)

Assertion

Ref	Expression
alimp-no-surprise	⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)

Proof of Theorem alimp-no-surprise

Step	Hyp	Ref	Expression
1		pm4.82 1021	. . . . 5 ⊢ (((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ ¬ 𝜑)
2	1	albii 1822	. . . 4 ⊢ (∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ ∀𝑥 ¬ 𝜑)
3		alnex 1784	. . . 4 ⊢ (∀𝑥 ¬ 𝜑 ↔ ¬ ∃𝑥𝜑)
4	2, 3	sylbb 218	. . 3 ⊢ (∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) → ¬ ∃𝑥𝜑)
5		imnan 400	. . 3 ⊢ ((∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) → ¬ ∃𝑥𝜑) ↔ ¬ (∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑))
6	4, 5	mpbi 229	. 2 ⊢ ¬ (∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑)
7		19.26 1873	. . . 4 ⊢ (∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)))
8	7	anbi2ci 625	. . 3 ⊢ ((∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑) ↔ (∃𝑥𝜑 ∧ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))))
9		3anass 1094	. . 3 ⊢ ((∃𝑥𝜑 ∧ ∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∃𝑥𝜑 ∧ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓))))
10		3anrot 1099	. . 3 ⊢ ((∃𝑥𝜑 ∧ ∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
11	8, 9, 10	3bitr2i 299	. 2 ⊢ ((∀𝑥((𝜑 → 𝜓) ∧ (𝜑 → ¬ 𝜓)) ∧ ∃𝑥𝜑) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
12	6, 11	mtbi 322	1 ⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   ∧ w3a 1086  ∀wal 1537  ∃wex 1782

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812

This theorem depends on definitions:  df-bi 206  df-an 397  df-3an 1088  df-ex 1783

This theorem is referenced by:  alsi-no-surprise  46500