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Proof step 9 from proof test-01 (previous step)

"15" ≠ "" ⊦ 21 = hexchar("15".char1) * 16 ^ (n→i(# "15") - 1) + hex("15".rstring)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ 5 = hex("5")

Rule argument

Additional information

Used simplifier rules: hexchar1: ⊦ hexchar('1') = 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ # "" = 0;%str ≠ "" ⊦ # (* str) = # str.rstring + 1;