Theorem wess 2188

Description: Subset theorem for the well-ordering predicate. Exercise 4 of [TakeutiZaring] p. 31.

Assertion

Ref	Expression
wess	⊢ (A ⊆ B → (R We B → R We A))

Proof of Theorem wess

Step	Hyp	Ref	Expression
1		frss 2173	. . 3 ⊢ (A ⊆ B → (R Fr B → R Fr A))
2		soss 2140	. . 3 ⊢ (A ⊆ B → (R Or B → R Or A))
3	1, 2	anim12d 431	. 2 ⊢ (A ⊆ B → ((R Fr B ∧ R Or B) → (R Fr A ∧ R Or A)))
4		df-we 2186	. 2 ⊢ (R We B ↔ (R Fr B ∧ R Or B))
5		df-we 2186	. 2 ⊢ (R We A ↔ (R Fr A ∧ R Or A))
6	3, 4, 5	3imtr4g 426	1 ⊢ (A ⊆ B → (R We B → R We A))

Syntax hints:   → wi 2   ∧ wa 196   ⊆ wss 1487   Or wor 2059   Fr wfr 2061   We wwe 2062

This theorem is referenced by:  wefrc 2195  wereu 2197  trssord 2216  ordelord 2221

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-or 197  df-an 198  df-3an 583  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-po 2128  df-so 2138  df-fr 2169  df-we 2186

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