Basic expression manipulation with Symbolics.jl

General Usage
2

Hi all,
In relation to the initial post, would someone know how to further simplify expression:
image

There is likely a way to cancel the minus sign and to simplify T/T in term on the right, but I haven’t managed so far.

Here’s the code:

using Symbolics, Latexify
@variables t x  
@variables s(t,x)  u(t,x)  c(t,x)  μ(t,x)  ρ(t,x)
@variables qs(t,x) qu(t,x) qc(t,x)
@variables Qs
@variables T(t,x)   
@variables λ Dc
Dt, dx = Differential(t), Differential(x)
solve  = Symbolics.solve_for 
subs   = Symbolics.substitute;

# Define balances and axioms
fu  = ρ*Dt(u) ~ - dx(qu) 
fc  = ρ*Dt(c) ~ - dx(qc) 
fs  = ρ*Dt(s) ~ - dx(qs) + Qs
LTE = Dt(u) ~  T*Dt(s) + μ*Dt(c)
latexify([fu; fc; fs; LTE])

# Evaluate entropy production
Qs1  = solve( fs, Qs )
Qs1  = subs(Qs1, Dt(s)=>solve(LTE,Dt(s)))
Qs1  = subs(Qs1, Dt(u)=>solve(fu,Dt(u)))
Qs1  = subs(Qs1, Dt(c)=>solve(fc,Dt(c)))
Qs1  = subs(Qs1, qu=>T*qs + μ*qc)
Qs1  = simplify_fractions(simplify(expand_derivatives(Qs1)))
TQs1 = T*simplify(subs(Qs1, [qs=>-λ/T*dx(T), qc=>-Dc*dx(μ)]), expand=true)
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