BiImpDiImpDiSample

⊢ (f ⊃ g) ⊃ f ⊃ g BiImpDiImpDiSample

Proof:

1
true ⊃ true
2
(true ⊃ true)
3
(f ⊃ g) ∧ (true ⊃ true)
⊃ (f ; true) ⊃ (g ; true)
4
(f ⊃ g) ⊃ (f ; true) ⊃ (g ; true)
2, 3,Prop
5
(f ⊃ g) ⊃ f ⊃ g
4, def. of 

qed

The following instance of Axiom StateImpBi illustrates why it is not subsumed by Inference Rule BiGen:

⊢ ¬Q ⊃ ¬Q

Here Q is a propositional variable. We cannot use BiGen since ¬Q is not a theorem.

2024-08-03
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