AndChopA

⊢ (f ∧ f1) ⌢ g ⊃ f ⌢ g AndChopA

Proof:

1
f ∧ f1 ⊃ f
2
(f ∧ f1) ⌢ g ⊃ f ⌢ g

qed

2024-08-03
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