Theorem alseuals

Description: "All some one" implies "all some": requiring exactly one witness is stronger than requiring at least one. Any consequence of an allsome statement is therefore a consequence of the corresponding "all some one" statement, which is how alseu-no-surprise is proved. (Contributed by David A. Wheeler, 21-Jul-2026)

	Ref	Expression
Assertion	alseuals	$$⊢\forall \exists !x\left(\phi \to \psi \right)\to \forall \exists x\left(\phi \to \psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		euex	$$⊢\exists !x\phi \to \exists x\phi$$
2	1	anim2i	$$⊢\left(\forall x\left(\phi \to \psi \right)\wedge \exists !x\phi \right)\to \left(\forall x\left(\phi \to \psi \right)\wedge \exists x\phi \right)$$
3		df-alseu	$$⊢\forall \exists !x\left(\phi \to \psi \right)\leftrightarrow \left(\forall x\left(\phi \to \psi \right)\wedge \exists !x\phi \right)$$
4		df-als	$$⊢\forall \exists x\left(\phi \to \psi \right)\leftrightarrow \left(\forall x\left(\phi \to \psi \right)\wedge \exists x\phi \right)$$
5	2 3 4	3imtr4i	$$⊢\forall \exists !x\left(\phi \to \psi \right)\to \forall \exists x\left(\phi \to \psi \right)$$