Metamath Proof Explorer


Theorem tposss

Description: Subset theorem for transposition. (Contributed by Mario Carneiro, 10-Sep-2015)

Ref Expression
Assertion tposss ( 𝐹𝐺 → tpos 𝐹 ⊆ tpos 𝐺 )

Proof

Step Hyp Ref Expression
1 coss1 ( 𝐹𝐺 → ( 𝐹 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ) ⊆ ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ) )
2 dmss ( 𝐹𝐺 → dom 𝐹 ⊆ dom 𝐺 )
3 cnvss ( dom 𝐹 ⊆ dom 𝐺 dom 𝐹 dom 𝐺 )
4 unss1 ( dom 𝐹 dom 𝐺 → ( dom 𝐹 ∪ { ∅ } ) ⊆ ( dom 𝐺 ∪ { ∅ } ) )
5 resmpt ( ( dom 𝐹 ∪ { ∅ } ) ⊆ ( dom 𝐺 ∪ { ∅ } ) → ( ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) ↾ ( dom 𝐹 ∪ { ∅ } ) ) = ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) )
6 2 3 4 5 4syl ( 𝐹𝐺 → ( ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) ↾ ( dom 𝐹 ∪ { ∅ } ) ) = ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) )
7 resss ( ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) ↾ ( dom 𝐹 ∪ { ∅ } ) ) ⊆ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } )
8 6 7 eqsstrrdi ( 𝐹𝐺 → ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ⊆ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) )
9 coss2 ( ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ⊆ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) → ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ) ⊆ ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) ) )
10 8 9 syl ( 𝐹𝐺 → ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ) ⊆ ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) ) )
11 1 10 sstrd ( 𝐹𝐺 → ( 𝐹 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) ) ⊆ ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) ) )
12 df-tpos tpos 𝐹 = ( 𝐹 ∘ ( 𝑥 ∈ ( dom 𝐹 ∪ { ∅ } ) ↦ { 𝑥 } ) )
13 df-tpos tpos 𝐺 = ( 𝐺 ∘ ( 𝑥 ∈ ( dom 𝐺 ∪ { ∅ } ) ↦ { 𝑥 } ) )
14 11 12 13 3sstr4g ( 𝐹𝐺 → tpos 𝐹 ⊆ tpos 𝐺 )