Hi, I have these two proofs that I can't wrap my head around. Each proof has two given premises, then a conclusion. The premises are divided from the conclusion by a dashed line.

Notes on syntax:
/\ : Conjunction. "x /\ y" means "x and y"
⊃: Material implication. "x ⊃ y" means "if x, then y"
≡: Material equivalence: "x ≡ y" means "x if and only if y"

(1)
1. (∀x)Sxx
2. (∀x)(∀y)(∀z)((Sxy /\ Sxz) ⊃ Syz)
--------------------------------
(∀x)(∀y)(Sxy ≡ Syx)

(2)
1. ∃y(∀x)(Ax ⊃ Byx)
2. ~∃z(Cz /\ Byz)
-------------------------------
~∃x(Cx /\ Ax)

If you'd like to solve them, I'm paying 500 fg each problem. Or if you have a hint, I'd be glad to learn more and attempt to solve the problems.

This post was edited by Darkemperor121 on Aug 16 2016 03:34pm

(1) Let x and y.

Suppose Sxy.
From premise 1, Sxx.
From premise 2, since Sxy and Sxx, then Syx.
Hence : Sxy ⊃ Syx.

Suppose Syx.
From premise 1, Syy.
From premise 2, since Syx and Syy, then Sxy.
Hence : Syx ⊃ Sxy.

Conclusion : Sxy ≡ Syx
And this is true for any x and y : (∀x)(∀y)(Sxy ≡ Syx)

(2) Suppose that : ∃x(Cx /\ Ax)
Denote that element x₀.

From premise 1, there exits y₀ such that Ax₀ implies By₀x₀
Since Ax₀ is true, then By₀x₀ is also true.

Hence, Cx₀ and By₀x₀.

Let's use premise 2, that should be : (∀y)(~∃z)(Cz /\ Byz).
We have found y₀ and x₀ such that Cx₀ /\ By₀x₀
This contradicts premise 2.

Conclusion : ~ ∃x(Cx /\ Ax)

This post was edited by feanur on Aug 17 2016 01:48pm

I don't see how this is valid. For Premise 2 we have:
Original: (Sxy /\ Sxz) ⊃ Syz
Rewritten: (Sxx /\ Sxy) ⊃ Sxy

This is different from what you wrote: (Syx /\ Syy) ⊃ Sxy

The solution for (1) doesn't seem right to me, but I'm sending 500 fg for solution 2.

Solution 1 attempts will stay open for another 40 minutes! (This is due at 8:00 PM EST, and I want to have an hour or so to finalize and submit everything)

This post was edited by Darkemperor121 on Aug 17 2016 04:25pm

Closed

Thanks for the tip.

What I wrote is correct however. Your premise 2 :
(∀x)(∀y)(∀z)((Sxy /\ Sxz) ⊃ Syz)
is true whatever x, y and z are (for all x, y, z).
Just take :
y for x
x for y
y for z
and rewrite :
(Syx /\ Syy) ⊃ Sxy