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Description: The rank of an ordered pair. Part of Exercise 4 of Kunen p. 107. (Contributed by Mario Carneiro, 10-Jun-2013)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | rankopb | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ 〈 𝐴 , 𝐵 〉 ) = suc suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | dfopg | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → 〈 𝐴 , 𝐵 〉 = { { 𝐴 } , { 𝐴 , 𝐵 } } ) | |
| 2 | 1 | fveq2d | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ 〈 𝐴 , 𝐵 〉 ) = ( rank ‘ { { 𝐴 } , { 𝐴 , 𝐵 } } ) ) |
| 3 | snwf | ⊢ ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) → { 𝐴 } ∈ ∪ ( 𝑅1 “ On ) ) | |
| 4 | prwf | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → { 𝐴 , 𝐵 } ∈ ∪ ( 𝑅1 “ On ) ) | |
| 5 | rankprb | ⊢ ( ( { 𝐴 } ∈ ∪ ( 𝑅1 “ On ) ∧ { 𝐴 , 𝐵 } ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ { { 𝐴 } , { 𝐴 , 𝐵 } } ) = suc ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) ) | |
| 6 | 3 4 5 | syl2an2r | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ { { 𝐴 } , { 𝐴 , 𝐵 } } ) = suc ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) ) |
| 7 | snsspr1 | ⊢ { 𝐴 } ⊆ { 𝐴 , 𝐵 } | |
| 8 | ssequn1 | ⊢ ( { 𝐴 } ⊆ { 𝐴 , 𝐵 } ↔ ( { 𝐴 } ∪ { 𝐴 , 𝐵 } ) = { 𝐴 , 𝐵 } ) | |
| 9 | 7 8 | mpbi | ⊢ ( { 𝐴 } ∪ { 𝐴 , 𝐵 } ) = { 𝐴 , 𝐵 } |
| 10 | 9 | fveq2i | ⊢ ( rank ‘ ( { 𝐴 } ∪ { 𝐴 , 𝐵 } ) ) = ( rank ‘ { 𝐴 , 𝐵 } ) |
| 11 | rankunb | ⊢ ( ( { 𝐴 } ∈ ∪ ( 𝑅1 “ On ) ∧ { 𝐴 , 𝐵 } ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ ( { 𝐴 } ∪ { 𝐴 , 𝐵 } ) ) = ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) ) | |
| 12 | 3 4 11 | syl2an2r | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ ( { 𝐴 } ∪ { 𝐴 , 𝐵 } ) ) = ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) ) |
| 13 | rankprb | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ { 𝐴 , 𝐵 } ) = suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) ) | |
| 14 | 10 12 13 | 3eqtr3a | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) = suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) ) |
| 15 | suceq | ⊢ ( ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) = suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) → suc ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) = suc suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) ) | |
| 16 | 14 15 | syl | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → suc ( ( rank ‘ { 𝐴 } ) ∪ ( rank ‘ { 𝐴 , 𝐵 } ) ) = suc suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) ) |
| 17 | 2 6 16 | 3eqtrd | ⊢ ( ( 𝐴 ∈ ∪ ( 𝑅1 “ On ) ∧ 𝐵 ∈ ∪ ( 𝑅1 “ On ) ) → ( rank ‘ 〈 𝐴 , 𝐵 〉 ) = suc suc ( ( rank ‘ 𝐴 ) ∪ ( rank ‘ 𝐵 ) ) ) |