Last time Sequent: 1 , . . . , n assumptions goal - PowerPoint PPT Presentation

Last time Sequent: φ 1 , . . . , φ n ⊢ ψ � �� � ���� assumptions goal if you can prove this and also this (using this rule) you can conclude this · · · ⊢ φ · · · ⊢ ψ ( ∧ introduction) · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ ( ∧ elimination) ( ∧ elimination) · · · ⊢ φ · · · ⊢ ψ · · · , φ ⊢ ψ · · · ⊢ φ · · · ⊢ φ → ψ · · · ⊢ φ → ψ ( → intro) ( → elim) · · · ⊢ ψ

· · · ⊢ φ · · · ⊢ ψ ( ∧ introduction) · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ ( ∧ elimination) ( ∧ elimination) · · · ⊢ φ · · · ⊢ ψ · · · , φ ⊢ ψ · · · ⊢ φ · · · ⊢ φ → ψ · · · ⊢ φ → ψ ( → intro) ( → elim) · · · ⊢ ψ

· · · ⊢ φ · · · ⊢ ψ ( ∧ introduction) · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ ( ∧ elimination) ( ∧ elimination) · · · ⊢ φ · · · ⊢ ψ · · · , φ ⊢ ψ · · · ⊢ φ · · · ⊢ φ → ψ · · · ⊢ φ → ψ ( → intro) ( → elim) · · · ⊢ ψ · · · ⊢ φ · · · ⊢ ψ · · · ⊢ φ ∨ ψ ( ∨ intro) · · · ⊢ φ ∨ ψ ( ∨ intro)

· · · ⊢ φ · · · ⊢ ψ ( ∧ introduction) · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ · · · ⊢ φ ∧ ψ ( ∧ elimination) ( ∧ elimination) · · · ⊢ φ · · · ⊢ ψ · · · , φ ⊢ ψ · · · ⊢ φ · · · ⊢ φ → ψ · · · ⊢ φ → ψ ( → intro) ( → elim) · · · ⊢ ψ · · · ⊢ φ · · · ⊢ ψ · · · ⊢ φ ∨ ψ ( ∨ intro) · · · ⊢ φ ∨ ψ ( ∨ intro) · · · ⊢ φ 1 ∨ φ 2 · · · , φ 1 ⊢ ψ · · · , φ 1 ⊢ ψ ( ∨ elim) · · · ⊢ ψ

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.

Claim: P ∨ ( Q ∧ R ) → (( P ∨ Q ) ∧ ( P ∨ R )) Proof: Assume P ∨ ( Q ∧ R ); we want to show ( P ∨ Q ) ∧ ( P ∨ R ) First, we will show P ∨ Q , then we will show P ∨ R . Since we know P ∨ ( Q ∧ R ), (by assumption) we can do case analysis. In the case where P is true, we have P ∨ Q , since we have P . We also have P ∨ R , so we can conclude ( P ∨ Q ) ∧ ( P ∨ R ). In the case where ( Q ∧ R ) is true, we again want to show ( P ∨ R ) ∧ ( Q ∨ R ). Since we know ( Q ∧ R ), we can conclude Q , and therefore P ∨ Q Similarly, since we have Q ∧ R we can conclude R . and therefore P ∨ R . Putting these together gives ( P ∨ Q ) ∧ ( P ∨ R ), as required.