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

Last time

Sequent: φ1, . . . , φn

• assumptions

⊢ ψ

• goal

if you can prove this and also this you can conclude this (using this rule) · · · ⊢ φ · · · ⊢ ψ · · · ⊢ φ ∧ ψ (∧ introduction) · · · ⊢ φ ∧ ψ · · · ⊢ φ (∧ elimination) · · · ⊢ φ ∧ ψ · · · ⊢ ψ (∧ elimination) · · · , φ ⊢ ψ · · · ⊢ φ → ψ (→ intro) · · · ⊢ φ · · · ⊢ φ → ψ · · · ⊢ ψ (→ elim)

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

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

· · · ⊢ φ · · · ⊢ ψ · · · ⊢ φ ∧ ψ (∧ introduction) · · · ⊢ φ ∧ ψ · · · ⊢ φ (∧ elimination) · · · ⊢ φ ∧ ψ · · · ⊢ ψ (∧ elimination) · · · , φ ⊢ ψ · · · ⊢ φ → ψ (→ intro) · · · ⊢ φ · · · ⊢ φ → ψ · · · ⊢ ψ (→ elim) · · · ⊢ φ · · · ⊢ φ ∨ ψ (∨ intro) · · · ⊢ ψ · · · ⊢ φ ∨ ψ (∨ intro) · · · ⊢ φ1 ∨ φ2 · · · , φ1 ⊢ ψ · · · , φ1 ⊢ ψ · · · ⊢ ψ (∨ elim)

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.

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.

· · · , φ ⊢ φ assumption

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof:

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P.

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P. We know that P is either true or false,

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P. We know that P is either true or false, · · · ⊢ φ ∨ ¬φ (law of excluded middle)

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P. We know that P is either true or false, so we can use case analysis: show P is true in either case. · · · ⊢ φ ∨ ¬φ (law of excluded middle)

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P. We know that P is either true or false, so we can use case analysis: show P is true in either case. If P is true, we are done. · · · ⊢ φ ∨ ¬φ (law of excluded middle)

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P. We know that P is either true or false, so we can use case analysis: show P is true in either case. If P is true, we are done. If P is false, we have a both ¬P and ¬(¬P), a contradiction. · · · ⊢ φ ∨ ¬φ (law of excluded middle)

· · · , φ ⊢ φ assumption Claim: ¬¬P → P Proof: Assume ¬¬P. We know that P is either true or false, so we can use case analysis: show P is true in either case. If P is true, we are done. If P is false, we have a both ¬P and ¬(¬P), a contradiction. · · · ⊢ φ ∨ ¬φ (law of excluded middle) · · · ⊢ φ · · · ⊢ ¬φ · · · ⊢ ψ (absurd)

φ1, . . . , φn ⊢ ψ: ◮ ψ is provable from the φi ◮ there is a valid proof tree ending in φ1, . . . , φn ⊢ ψ

φ1, . . . , φn ⊢ ψ: ◮ ψ is provable from the φi ◮ there is a valid proof tree ending in φ1, . . . , φn ⊢ ψ φ1, . . . , φn ψ: ◮ the φi entail ψ ◮ for all I that satisfy all of the φi, I also satisfies ψ.

φ1, . . . , φn ⊢ ψ: ◮ ψ is provable from the φi ◮ there is a valid proof tree ending in φ1, . . . , φn ⊢ ψ φ1, . . . , φn ψ: ◮ the φi entail ψ ◮ for all I that satisfy all of the φi, I also satisfies ψ. Defn: ⊢ is sound (with respect to ) if, whenever (φi) ⊢ ψ, we have (φi) ψ. Defn: ⊢ is complete (with respect to ) if, whenever (φi) ψ, we have (φi) ⊢ ψ.