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Your test bool(eq1[0][0].rhs()>eq2[0][0].rhs()) doesn't make any hypothesis onx_1. But :

sage: solve(eq1[0].rhs()>eq2[0].rhs(),x)
[[x_1 > -58]]

Since there are values of x_1 for which your predicate is false, bool is right to return False. Checking other cases is left as an exercise to the reader.

HTH,

PS : your notations could be somewhat streamlined. Learning a bit of Python would be a very wise investment...