Theorem alsi-no-surprise 46500

Description: Demonstrate that there is never a "surprise" when using the allsome quantifier, that is, it is never possible for the consequent to be both always true and always false. This uses the definition of df-alsi 46492; the proof itself builds on alimp-no-surprise 46485. For a contrast, see alimp-surprise 46484. (Contributed by David A. Wheeler, 27-Oct-2018.)

Assertion

Ref	Expression
alsi-no-surprise	⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))

Proof of Theorem alsi-no-surprise

Step	Hyp	Ref	Expression
1		alimp-no-surprise 46485	. 2 ⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
2		df-alsi 46492	. . . 4 ⊢ (∀!𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
3		df-alsi 46492	. . . 4 ⊢ (∀!𝑥(𝜑 → ¬ 𝜓) ↔ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
4	2, 3	anbi12i 627	. . 3 ⊢ ((∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)))
5		anandi3r 1102	. . 3 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)))
6		3ancomb 1098	. . 3 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
7	4, 5, 6	3bitr2i 299	. 2 ⊢ ((∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
8	1, 7	mtbir 323	1 ⊢ ¬ (∀!𝑥(𝜑 → 𝜓) ∧ ∀!𝑥(𝜑 → ¬ 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   ∧ w3a 1086  ∀wal 1537  ∃wex 1782  ∀!walsi 46490

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  df-alsi 46492

This theorem is referenced by: (None)