homework #6 solutions

LING 230 Homework #6: Natural Deduction with Quantification Fall 2009
solutions
Problem 1: Short logical arguments.
a. 1 ¬∃x(Px→Qx) assumption
2 ∀x¬(Px→Qx) QN, 1
3 ¬(Pg→Qg) E∀/UI, 2
4 ¬(¬Pg∨Qg) Impl, 3
5 ¬¬Pg∧¬Qg DeM, 4
6 ¬Qg E∧, 5
b. 1 ¬∃x∃y(Ay∧Ex) assumption
2 ∀x¬∃y(Ay∧Ex) QN, 1
3 ∀x∀y¬(Ay∧Ex) QN, 2
4 ∀y¬(Ay∧Ee) E∀/UI, 3
5 ¬(Aa∧Ee) E∀/UI, 4
6 ¬Aa∨¬Ee DeM, 5
c. 1 ∀x∀yRxy assumption
2 ∀yRay E∀/UI, 1
3 Rab E∀/UI, 2
4 ∀yRby E∀/UI, 1
5 Rba E∀/UI, 4
6 Rba∧Rab I∧, 5, 3
7 ∃x(Rbx∧Rxb I∃/EG, 6
8 ∃y∃x(Ryx∧Rxy) I∃/EG, 7
d. 1 ¬∃z(¬Gz∨Gz) assumption
2 ∀z¬(¬Gz∨Gz) QN, 1
3 ¬(¬Ga∨Ga) E∀/UI, 2
4 ¬¬Ga∧¬Ga DeM, 3
5 ¬¬Ga E∧, 4
6 ¬Ga E∧, 4
7 ⊥ E¬, 5, 6
8 ¬¬∃z(¬Gz∨Gz) I¬/RAA
9 ∃z(¬Gz∨Gz) DN, 8
Problem 2: Partially completed longer logical arguments.
1 ¬∃xPx assumption
2 Pa assumption
3 ∃xPx I∃/EG, 2
4 ⊥ E¬, 1, 3
5 ¬Pa I¬/RAA
6 ∀x¬Px I∀/UG, 5
1 ∀x¬Px assumption
2 ∃xPx assumption
3 Pa assumption
4 ¬Pa E∀/UI, 1
5 ⊥ E¬, 4, 3
6 ¬∀x¬Px EFSQ, 5
7 Pa→¬∀x¬Px I→/CP
8 ¬∀x¬Px E∃/EI, 2, 7
9 ⊥ E¬, 8, 1
10 ¬∃xPx I¬/RAA
1 ¬∀xPx assumption
2 ¬∃x¬Px assumption
3 ¬Pa assumption
4 ∃x¬Px I∃/EG, 3
5 ⊥ E¬, 2, 4
6 ¬¬Pa I¬/RAA
7 Pa DN, 6
8 ∀xPx I∀/UG, 7
9 ⊥ E¬, 1, 8
10 ¬¬∃x¬Px I¬/RAA
11 ∃x¬Px DN, 10
1 ∃x¬Px assumption
2 ¬Pa assumption
3 ∀xPx assumption
4 Pa E∀/UI, 3
5 ⊥ E¬, 2, 4
6 ¬∀xPx I¬/RAA
7 ¬Pa→¬∀xPx I→/CP
8 ¬∀xPx E∃/EI, 1, 7

Problem 3: Longer longer logical arguments.
a. 1 ∀x(Ax→Bx) assumption
2 ∀x(Bx→ Cx) assumption
3 ∀xAx assumption
4 A j → B j E∀/UI, 1
5 B j → C j E∀/UI, 2
6 A j E∀/UI, 3
7 B j E→/MP, 4, 6
8 C j E→/MP, 5, 7
9 B j∧C j I∧, 7, 8
10 ∀x(Bx∧Cx) I∀/UG, 9
b. 1 ∀x(Ax→Bx) assumption
2 Ag→Bg E∀/UI, 1
3 Ah→Bh E∀/UI, 1
4 Ag∧Ah assumption
5 Ag E∧, 4
6 Ah E∧, 4
7 Bg E→/MP, 2, 5
8 Bh E→/MP, 3, 6
9 Bg∧Bh I∧, 7, 8
10 (Ag∧Ah)→(Bg∧Bh) I→/CP
11 ∀y ( (Ag∧Ay)→(Bg∧By) ) I∀/UG, 10
12 ∀x∀y ( (Ax∧Ay)→(Bx∧By) ) I∀/UG, 11
c. 1 ∃x(Px∨Qx) assumption
2 Pa∨Qa assumption
3 Pa assumption
4 ∃yPy I∃/EG, 3
5 Pa→∃yPy I→/CP
6 Qa assumption
7 ∃zQz I∃/EG, 6
8 Qa→∃zQz I→/CP
9 ∃yPy∨∃zQz CD, 2, 5, 8
10 (Pa∨Qa)→(∃yPy∨∃zQz) I→/CP
11 ∃yPy∨∃zQz E∃/EI, 1, 10
d. 1 ¬∀z(¬Gz∨Gz) assumption
2 ∃z¬(¬Gz∨Gz) QN, 1
3 ¬(¬Gl∨Gl) assumption
4 ¬¬Gl∧¬Gl DeM, 3
5 ¬¬Gl E∧, 4
6 ¬Gl E∧, 4
7 ⊥ E¬, 5, 6
8 ∀z(¬Gz∨Gz) EFSQ, 7
9 ¬(¬Gl∨Gl)→∀z(¬Gz∨Gz) I→/CP
10 ∀z(¬Gz∨Gz) E∃/EI, 2, 9
11 ⊥ E¬, 1, 10
12 ¬¬∀z(¬Gz∨Gz) I¬/RAA
13 ∀z(¬Gz∨Gz) DN, 12