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BfChopEqvChop
⊢
(
f
≡
f
1
)
⊃
(
f ⌢ g
)
≡
(
f
1
⌢ g
)
BfChopEqvChop
Proof:
1
(
f
≡
f
1
)
≡
(
f
⊃
f
1
)
∧
(
f
1
⊃
f
)
BfEqvSplit
2
(
f
⊃
f
1
)
⊃
(
f ⌢ g
)
⊃
(
f
1
⌢ g
)
BfChopImpChop
3
(
f
1
⊃
f
)
⊃
(
f
1
⌢ g
)
⊃
(
f ⌢ g
)
BfChopImpChop
4
(
f
≡
f
1
)
⊃
(
f ⌢ g
)
≡
(
f
1
⌢ g
)
1
−−
3
,
Prop
qed
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2023-09-12
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