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Proof step 111 from proof byte-law-righthift-not (previous step)

8 ≠ 0, i2b(bits2uint(false ' + firstn(8 -1, true ' + false ' + false ' + false ' + false ' + false ' + false ' + false '))) = i2b(2147483392 + bits2uint(true ' + firstn(7, true ' + false ' + false ' + false ' + false ' + false ' + false ' + false '))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits2uint(false ' + bits) = bits2uint(bits);