homework #6 solutions

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