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AndChopAndCommute
⊢
(
f
∧
g
) ; (
f
1
∧
g
1
)
≡
(
g
∧
f
) ; (
g
1
∧
f
1
)
AndChopAndCommute
Proof:
1
⊢
(
f
∧
g
) ; (
f
1
∧
g
1
)
≡
(
g
∧
f
) ; (
f
1
∧
g
1
)
AndChopCommute
2
⊢
(
g
∧
f
) ; (
f
1
∧
g
1
)
≡
(
g
∧
f
) ; (
g
1
∧
f
1
)
ChopAndCommute
3
⊢
(
f
∧
g
) ; (
f
1
∧
g
1
)
≡
(
g
∧
f
) ; (
g
1
∧
f
1
)
1
,
2
,
EqvChain
qed
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2023-09-12
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