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Mirrors > Home > Home > Th. List > ge-go |
Description: Conjunction implies biimplication. (Contributed by la korvo, 25-Jun-2024.) |
Ref | Expression |
---|---|
ge-go.0 | ⊢ ge broda gi brode |
Ref | Expression |
---|---|
ge-go | ⊢ go broda gi brode |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ge-go.0 | . . 3 ⊢ ge broda gi brode | |
2 | ge-ganai 51 | . . 3 ⊢ ganai ge broda gi brode gi ganai broda gi brode | |
3 | 1, 2 | ax-mp 10 | . 2 ⊢ ganai broda gi brode |
4 | ge-com-lem 119 | . . . 4 ⊢ ganai ge broda gi brode gi ge brode gi broda | |
5 | 1, 4 | ax-mp 10 | . . 3 ⊢ ge brode gi broda |
6 | ge-ganai 51 | . . 3 ⊢ ganai ge brode gi broda gi ganai brode gi broda | |
7 | 5, 6 | ax-mp 10 | . 2 ⊢ ganai brode gi broda |
8 | 3, 7 | iso 65 | 1 ⊢ go broda gi brode |
Colors of variables: sumti selbri bridi |
Syntax hints: ganai bgan 9 ge bge 42 |
This theorem was proved from axioms: ax-mp 10 ax-k 11 ax-s 15 ax-ge-le 43 ax-ge-re 44 ax-ge-in 45 |
This theorem depends on definitions: df-go 61 |
This theorem is referenced by: simsa-mintu 342 |
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