Theorem 3sstr3g 3199

Description: Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 1-Oct-2000.)

Hypotheses

Ref	Expression
3sstr3g.1	⊢ (𝜑 → 𝐴 ⊆ 𝐵)
3sstr3g.2	⊢ 𝐴 = 𝐶
3sstr3g.3	⊢ 𝐵 = 𝐷

Assertion

Ref	Expression
3sstr3g	⊢ (𝜑 → 𝐶 ⊆ 𝐷)

Proof of Theorem 3sstr3g

Step	Hyp	Ref	Expression
1		3sstr3g.1	. 2 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
2		3sstr3g.2	. . 3 ⊢ 𝐴 = 𝐶
3		3sstr3g.3	. . 3 ⊢ 𝐵 = 𝐷
4	2, 3	sseq12i 3185	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ 𝐶 ⊆ 𝐷)
5	1, 4	sylib 122	1 ⊢ (𝜑 → 𝐶 ⊆ 𝐷)

Syntax hints:   → wi 4   = wceq 1353   ⊆ wss 3131

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-11 1506  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-in 3137  df-ss 3144

This theorem is referenced by:  hmeontr  13852