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Proof step 12 from proof test-01 (previous step)

"5" ≠ "" ⊦ 5 = hexchar("5".char1) * 16 ^ (n→i(# "5") - 1) + hex("5".rstring)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: hex-empty: ⊦ hex("") = 0;%hexchar5: ⊦ hexchar('5') = 5;% ⊦ # "" = 0;%str ≠ "" ⊦ # (* str) = # str.rstring + 1;