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Proof step 3 from proof test-01 (previous step)

"9F15" ≠ "" ⊦ hexchar("9F15".char1) * 16 ^ (n→i(# "9F15") - 1) + hex("9F15".rstring) = 40725

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ 3861 = hex("F15")

Rule argument

Additional information

Used simplifier rules: hexchar9: ⊦ hexchar('9') = 9;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ # "" = 0;%str ≠ "" ⊦ # (* str) = # str.rstring + 1;