Theorem pssssd 1568

Description: Deduce subclass from proper subclass.

Hypothesis

Ref	Expression
pssssd.1	⊢ (φ → A ⊂ B)

Assertion

Ref	Expression
pssssd	⊢ (φ → A ⊆ B)

Proof of Theorem pssssd

Step	Hyp	Ref	Expression
1		pssssd.1	. 2 ⊢ (φ → A ⊂ B)
2		pssss 1567	. 2 ⊢ (A ⊂ B → A ⊆ B)
3	1, 2	syl 12	1 ⊢ (φ → A ⊆ B)

Syntax hints:   → wi 2   ⊆ wss 1487   ⊂ wpss 1488

This theorem is referenced by:  elprpq 3889  genpss 3901  ltexprlem7 3942

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198  df-pss 1494

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