Theorem 3sstr4d 1543

Description: Substitution of equality into both sides of a subclass relationship.

Hypotheses

Ref	Expression
3sstr4d.1	|- (ph -> A (_ B)
3sstr4d.2	|- (ph -> C = A)
3sstr4d.3	|- (ph -> D = B)

Assertion

Ref	Expression
3sstr4d	|- (ph -> C (_ D)

Proof of Theorem 3sstr4d

Step	Hyp	Ref	Expression
1		3sstr4d.1	. 2 |- (ph -> A (_ B)
2		3sstr4d.2	. . 3 |- (ph -> C = A)
3	2	cleqcomd 1106	. 2 |- (ph -> A = C)
4		3sstr4d.3	. . 3 |- (ph -> D = B)
5	4	cleqcomd 1106	. 2 |- (ph -> B = D)
6	1, 3, 5	3sstr3d 1542	1 |- (ph -> C (_ D)

Syntax hints:   -> wi 2   = wceq 1091   (_ wss 1487

This theorem is referenced by:  mapss 3270  occont 5168  spanss 5319

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492

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