Theorem nsstr 46031

Description: If it's not a subclass, it's not a subclass of a smaller one. (Contributed by Glauco Siliprandi, 26-Jun-2021.)

Assertion

Ref	Expression
nsstr	⊢ ((¬ 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐵) → ¬ 𝐴 ⊆ 𝐶)

Proof of Theorem nsstr

Step	Hyp	Ref	Expression
1		sstr 3938	. . . 4 ⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵) → 𝐴 ⊆ 𝐵)
2	1	ancoms 464	. . 3 ⊢ ((𝐶 ⊆ 𝐵 ∧ 𝐴 ⊆ 𝐶) → 𝐴 ⊆ 𝐵)
3	2	adantll 727	. 2 ⊢ (((¬ 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐵) ∧ 𝐴 ⊆ 𝐶) → 𝐴 ⊆ 𝐵)
4		simpll 779	. 2 ⊢ (((¬ 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐵) ∧ 𝐴 ⊆ 𝐶) → ¬ 𝐴 ⊆ 𝐵)
5	3, 4	pm2.65da 829	1 ⊢ ((¬ 𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐵) → ¬ 𝐴 ⊆ 𝐶)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ⊆ wss 3898

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-an 402  df-ss 3915

This theorem is used by:  mbfpsssmf  47715