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Proof step 2 from proof square-abs (previous step)

dec = mkDecimal(i, i0) ⊦ abs(dec * dec) = dec * dec

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ abs(mkDecimal(i, i0)) * abs(mkDecimal(i, i0)) = mkDecimal(i, i0) * mkDecimal(i, i0)

Rule argument

Additional information

Used simplifier rules: mult-abs: ⊦ abs(dec * dec0) = abs(dec) * abs(dec0);%mult-a: ⊦ dec0 * dec1 * dec2 = (dec0 * dec1) * dec2;%mult-c: ⊦ dec_i * dec_j = dec_j * dec_i;