ChopPlusImpChopPlus

f g ⇒ ⊢ f+ g+ ChopPlusImpChopPlus

Proof:

1
f g
given
2
f+ f (f more) ; f+
3
g+ g (g more) ; g+
4
f+ ∧¬(g+) ((f more) ; f+) ∧¬((g more) ; g+)
1, 2, 3,Prop
5
f more g more
1,Prop
6
(f more) ; f+ (g more) ; f+
7
f+ ∧¬(g+) ((g more) ; f+) ∧¬((g more) ; g+)
4, 6,Prop
8
g more more
9
f+ g+

qed

2024-08-03
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