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BfBoxEqvBoxBf
⊢
f
≡
f
BfBoxEqvBoxBf
Proof:
1
¬
f
≡
¬
f
DfDiamondEqvDiamondDf
2
¬
f
≡¬
f
DfDiamondNotEqvNotBfBox
3
¬
f
≡¬
f
DiamondDfNotEqvNotBoxBf
4
f
≡
f
1
−
3
,
Prop
qed
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2024-08-03
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