Proof Tactics, Strategies and Derived Rules. CS 270 Math Foundations of CS Jeremy Johnson
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1 Proof Tactics, Strategies and Derived Rules CS 270 Math Foundations of CS Jeremy Johnson
2 Outline 1. Review Rules 2. Positive subformulas and extraction 3. Proof tactics Extraction, Conversion, Inversion, Division, and Refutation Finding contradictions 4. Proof strategy Search tree and an algorithm to find a proof 5. Derived rules
3 Conjunction Rules Introduction Rule φ ψ φ ψ I Elimination Rule φ ψ φ ER φ ψ ψ EL
4 Implication Rules Introduction Rule φ ψ φ ψ I Assume φ and show ψ Elimination Rule (Modus Ponens) φ φ ψ ψ E
5 Disjunction Rules Introduction Rule φ φ ψ IR ψ φ ψ IL Elimination Rule (proof by case analysis) φ ψ φ χ χ ψ χ E
6 Negation Rules Introduce the symbol ( = bottom) to encode a contradiction Bottom elimination φ E Bottom introduction can prove anything φ φ I
7 Negation Rules Introduction and elimination rules φ φ I Double negation φ φ E φ φ E
8 Law of the Excluded Middle pp pp [derived rule LEM] 1 p p goal
9 Law of the Excluded Middle pp pp [derived rule LEM] 1 (p p) assumption 2 goal 3 p p E 2
10 Law of the Excluded Middle pp pp [derived rule LEM] 1 (p p) assumption 2 p p Goal 3 I 1,2 4 p p E 3
11 Law of the Excluded Middle pp pp [derived rule LEM] 1 (p p) assumption 2 p Goal 3 p p IR 2 4 I 1,3 5 p p E 4
12 Law of the Excluded Middle pp pp [derived rule LEM] 1 (p p) assumption 2 pp assumption 3 Goal 4 p E 3 5 p p IR 4 6 I 1,5 7 p p E 6
13 Law of the Excluded Middle pp pp [derived rule LEM] 1 (p p) assumption 2 pp assumption 3 (p p) IL 2 4 i 3,1 5 p E 4 6 p p IR 5 7 I 6,1 8 p p E 7
14 Search Tree? P? P (P P)? IL IR E? P P
15 Search Tree P? E? P? P (P P)? IL IR E? P P
16 P? P Search Tree I P? E The only possible contradictory pair is P and P and P? P can only be addressed by E and we are back where we started. backtrack? P IL
17 Search Tree? P? P (P P)? IL IR E? P P
18 Proof Tactics Systematically search for a proof Apply (,, ) elimination rules forward Apply introduction rules backwards No extraneous steps Backtrack when dead-end reached 1. Extraction 2. Conversion 3. Inversion 4. Division 5. Refutation
19 Positive Subformulas PS(ϕ) If ϕ is an atom return ϕ If ϕ = ψ return ϕ If ϕ = ψ ρ then return ϕ PS(ψ) PS(ρ) If ϕ = ψ ρ then return ϕ PS(ψ) PS(ρ) If ϕ = ψ ρ then return ϕ PS(ρ)
20 Extraction Apply elimination rules forward in order to extract goals that occur as positive subformulae of the formulae on available lines.
21 Conversion Use disjunction elimination in order to obtain goal disjunctions.
22 Inversion Invert non-atomic goals by applying introduction rules backward to them.
23 Division Use disjunction elimination on any goals for which the previous three tactics have either not applied, or not been successful.
24 Refutation Apply negation elimination backward to goals that cannot be obtained by any other means.
25 Possible Contradictions Form a list of all negations that appear as a positive subformulas of all premises and available assumptions. Pair each negation ϕ with its immediate subformula ϕ. These pairs are the only possible contradictions that must be considered.
26 Exercise Prove the definition of conditional (ϕ ψ) ϕ ψ ϕ ψ (ϕ ψ)
27 Deadend
28 Solution
29 Solution
30 Algorithm
31 Using Derived Rules Once you have proven a rule from the basic rules you may use it in your proofs Derive M from (M O) M 1 (M O) M premise 2 M assumption 3 M O IR 1 4 M O Df I 5 M E1,4 6 I2,5 7 M E 6
32 Derived Rules Commutative rules ϕ ψ ψ ϕ ϕ ψ ψ ϕ Associative rules (ϕ ψ) ρ (ψ ϕ) ρ (ϕ ψ) ρ (ψ ϕ) ρ Idempotence rules ϕ ϕ ϕ and ϕ ϕ ϕ ϕ ϕ ϕ and ϕ ϕ ϕ
33 Derived Rules Distributive rules ϕ (ψ ρ) (ϕ ψ) (ϕ ρ) (ϕ ψ) (ϕ ρ) ϕ (ψ ρ) ϕ (ψ ρ) (ϕ ψ) (ϕ ρ) (ϕ ψ) (ϕ ρ) ϕ (ψ ρ) Disjunctive syllogism (ϕ ψ), ψ ϕ Cut (resolution) ( ϕ ψ), ( ϕ ρ) (ψ ρ)
34 Derived Rules DeMorgan s rules (ϕ ψ) ϕ ψ ϕ ψ (ϕ ψ) (ϕ ψ) ϕ ψ ϕ ψ (ϕ ψ)
35 Derived Rules Modus Tollens (ϕ ψ, ψ) ϕ Transposition (ϕ ψ) ( ψ ϕ) Hypothetical Syllogism (ϕ ψ, ψ ρ) (ϕ ρ) Exportation and Importation ((ϕ ψ) ρ) (ϕ (ψ ρ)) (ϕ (ψ ρ)) ((ϕ ψ) ρ)
36 Derived Rules Definition of conditional (ϕ ψ) ϕ ψ ϕ ψ (ϕ ψ) Negated conditional (ϕ ψ) ϕ ψ ϕ ψ (ϕ ψ)
37 Exercise Prove the definition of the conditional using Disjunctive Syllogism and LEM
38 Solution 1 P Q premise 2 P assumption 3 Q DSL 1,2 4 P Q I3 1 P Q premise 2 P P LEM 3 P assumption 4 Q E1,3 5 P Q IL4 6 P assumption 7 P Q IR6 8 P Q E2,5,7
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