Theorem rals-no-surprise 50585

Description: Demonstrate that there is never a "surprise" when using the allsome quantifier restricted to a class, that is, it is never possible for the consequent to be both always true and always false of the members of 𝐴 that satisfy the antecedent. This is the restricted counterpart of als-no-surprise 50584, and follows from it by dfrals2 50568. Note that this needs no assumption that 𝐴 is nonempty, because allsome requires a member of 𝐴 satisfying 𝜑, and that member would have to satisfy both 𝜓 and ¬ 𝜓. The ordinary restricted "for all" requires no such member and can be vacuously true, as shown in empty-surprise2 50561; that is the point of allsome. (Contributed by David A. Wheeler, 12-Jul-2026.)

Assertion

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

Proof of Theorem rals-no-surprise

Step	Hyp	Ref	Expression
1		als-no-surprise 50584	. 2 ⊢ ¬ (∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓) ∧ ∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → ¬ 𝜓))
2		dfrals2 50568	. . 3 ⊢ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ↔ ∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓))
3		dfrals2 50568	. . 3 ⊢ (∀∃𝑥 ∈ 𝐴(𝜑 → ¬ 𝜓) ↔ ∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → ¬ 𝜓))
4	2, 3	anbi12i 639	. 2 ⊢ ((∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∀∃𝑥 ∈ 𝐴(𝜑 → ¬ 𝜓)) ↔ (∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝜓) ∧ ∀∃𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → ¬ 𝜓)))
5	1, 4	mtbir 326	1 ⊢ ¬ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∀∃𝑥 ∈ 𝐴(𝜑 → ¬ 𝜓))

Syntax hints:  ¬ wn 3   ∧ wa 400   ∈ wcel 2143  ∀∃wals 50564  ∀∃wrals 50565

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839

This theorem depends on definitions:  df-bi 210  df-an 401  df-3an 1105  df-ex 1810  df-ral 3080  df-rex 3090  df-als 50566  df-rals 50567

This theorem is referenced by: (None)