Theorem als-no-surprise 50464

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-als 50446; the proof itself builds on alimp-no-surprise 50439. For a contrast, see alimp-surprise 50438. (Contributed by David A. Wheeler, 27-Oct-2018.)

Assertion

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

Proof of Theorem als-no-surprise

Step	Hyp	Ref	Expression
1		alimp-no-surprise 50439	. 2 ⊢ ¬ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)
2		df-als 50446	. . . 4 ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
3		df-als 50446	. . . 4 ⊢ (∀∃𝑥(𝜑 → ¬ 𝜓) ↔ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
4	2, 3	anbi12i 639	. . 3 ⊢ ((∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)))
5		anandi3r 1118	. . 3 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ∧ (∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑)))
6		3ancomb 1114	. . 3 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑 ∧ ∀𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
7	4, 5, 6	3bitr2i 302	. 2 ⊢ ((∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓)) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜑 → ¬ 𝜓) ∧ ∃𝑥𝜑))
8	1, 7	mtbir 326	1 ⊢ ¬ (∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 400   ∧ w3a 1101  ∀wal 1565  ∃wex 1806  ∀∃wals 50444

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836

This theorem depends on definitions:  df-bi 210  df-an 401  df-3an 1103  df-ex 1807  df-als 50446

This theorem is referenced by:  rals-no-surprise  50465