Axiomatizing Probabilistic Logic of Quantum Programs
Jort Bergfeld and Joshua Sack Amsterdam, 2014 April 1
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Axiomatizing Probabilistic Logic of Quantum Programs Jort Bergfeld - - PowerPoint PPT Presentation
Axiomatizing Probabilistic Logic of Quantum Programs Jort Bergfeld and Joshua Sack Amsterdam, 2014 April 1 1/32 Motivation and Background Quantum Algorithms and Protocols: use logic for a better understanding. Probabilistic Logic of Quantum
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1 there exists an i ∈ N such that 1
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2 |
3
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1 there exists an i ∈ N such that 1
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2 |f (bk)|2 ∈ [0, 1], and 3
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1 φ, φ → ψ =
2 φ =
3 φ → [p?]ψ =
4 φ =
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1 p → (⊤ → p?⊤) 2 p?⊤ ↔ ♦p 3 p → p?⊤ 4 p → ♦p 21/32
1 p → (⊤ → p?⊤) 2 p?⊤ ↔ ♦p 3 p → p?⊤ 4 p → ♦p 21/32
1 p → (⊤ → p?⊤) 2 p?⊤ ↔ ♦p 3 p → p?⊤ 4 p → ♦p 21/32
1 p → (⊤ → p?⊤) 2 p?⊤ ↔ ♦p 3 p → p?⊤ 4 p → ♦p 21/32
1 p → ¬∼p 2 (p ∧ ∼p) → (¬∼p ∧ ∼p) 3 (p ∧ ∼p) → ⊥ 4 ¬(p ∧ ∼p) 5 ¬(p ∧ ∼p) 6 ∼(p ∧ ∼p) 22/32
1 p → ¬∼p 2 (p ∧ ∼p) → (¬∼p ∧ ∼p) 3 (p ∧ ∼p) → ⊥ 4 ¬(p ∧ ∼p) 5 ¬(p ∧ ∼p) 6 ∼(p ∧ ∼p) 22/32
1 p → ¬∼p 2 (p ∧ ∼p) → (¬∼p ∧ ∼p) 3 (p ∧ ∼p) → ⊥ 4 ¬(p ∧ ∼p) 5 ¬(p ∧ ∼p) 6 ∼(p ∧ ∼p) 22/32
1 p → ¬∼p 2 (p ∧ ∼p) → (¬∼p ∧ ∼p) 3 (p ∧ ∼p) → ⊥ 4 ¬(p ∧ ∼p) 5 ¬(p ∧ ∼p) 6 ∼(p ∧ ∼p) 22/32
1 p → ¬∼p 2 (p ∧ ∼p) → (¬∼p ∧ ∼p) 3 (p ∧ ∼p) → ⊥ 4 ¬(p ∧ ∼p) 5 ¬(p ∧ ∼p) 6 ∼(p ∧ ∼p) 22/32
1 p → ¬∼p 2 (p ∧ ∼p) → (¬∼p ∧ ∼p) 3 (p ∧ ∼p) → ⊥ 4 ¬(p ∧ ∼p) 5 ¬(p ∧ ∼p) 6 ∼(p ∧ ∼p) 22/32
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1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
1 ⊤ → (p → ⊤?p) 2 p → ⊤?p 3 ⊤?p → [⊤?]p 4 p → [⊤?]p 5 [⊤?]p → p 6 ⊤?p → p 7 p → [⊤?]⊤?♦p 8 p → ♦p 24/32
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1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
1 p → ♦p
2 p → ♦p
3 T(∼p)
4 p → p ⊔ q 5 (p → q) → (∼q → ∼p) 6 T(p) ∧ T(q) → T(p ∧ q) 7 p ⊥ q ↔ q ⊥ p 8 r ⊥ p ∧ r ⊥ q ↔ r ⊥ (p ⊔ q) 26/32
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1 (Pr(p) = 0) ↔ ∼p 2 (Pr(p) > 0) ↔ ♦p 3 p → (Pr(p) = 1) 4 T(p) → (p ↔ Pr(p) = 1) 5 Orthocomplement
6 Quantum join
7 Superposition
8 (p ≤ q) ∧ q?=ρ(Pr(p) = τ) → (Pr(p) = ρτ) 28/32
1 (Pr(p) = 0) ↔ ∼p 2 (Pr(p) > 0) ↔ ♦p 3 p → (Pr(p) = 1) 4 T(p) → (p ↔ Pr(p) = 1) 5 Orthocomplement
6 Quantum join
7 Superposition
8 (p ≤ q) ∧ q?=ρ(Pr(p) = τ) → (Pr(p) = ρτ) 28/32
1 (Pr(p) = 0) ↔ ∼p 2 (Pr(p) > 0) ↔ ♦p 3 p → (Pr(p) = 1) 4 T(p) → (p ↔ Pr(p) = 1) 5 Orthocomplement
6 Quantum join
7 Superposition
8 (p ≤ q) ∧ q?=ρ(Pr(p) = τ) → (Pr(p) = ρτ) 28/32
1 (Pr(p) = 0) ↔ ∼p 2 (Pr(p) > 0) ↔ ♦p 3 p → (Pr(p) = 1) 4 T(p) → (p ↔ Pr(p) = 1) 5 Orthocomplement
6 Quantum join
7 Superposition
8 (p ≤ q) ∧ q?=ρ(Pr(p) = τ) → (Pr(p) = ρτ) 28/32
1 (Pr(p) = 0) ↔ ∼p 2 (Pr(p) > 0) ↔ ♦p 3 p → (Pr(p) = 1) 4 T(p) → (p ↔ Pr(p) = 1) 5 Orthocomplement
6 Quantum join
7 Superposition
8 (p ≤ q) ∧ q?=ρ(Pr(p) = τ) → (Pr(p) = ρτ) 28/32
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
1 Ep1 ∧ Ep2 ∧ (p1 ⊥ p2) → E
3 T(p1 ⊔ p2) → (Pr(p1 ⊔ p2) = 1) → p1 ⊔ p2 4 (p3 ⊥ p1) ∧ (p3 ⊥ p2) → p3 ⊥ p1 ⊔ p2 5
3 ((Pr(p1) = 1
7 E
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