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AndChopImpChopAndChop
⊢
(
f
∧
f
1
)
⌢ g
⊃
(
f ⌢ g
)
∧
(
f
1
⌢ g
)
AndChopImpChopAndChop
Proof:
1
(
f
∧
f
1
)
⌢ g
⊃
f ⌢ g
AndChopA
2
(
f
∧
f
1
)
⌢ g
⊃
f
1
⌢ g
AndChopB
3
(
f
∧
f
1
)
⌢ g
⊃
(
f ⌢ g
)
∧
(
f
1
⌢ g
)
1
,
2
,
Prop
qed
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2023-09-12
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