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Proof step 12 from proof bits2int-sign (previous step)

(bits0 + true ' + bits1).first ∧ bits0 + true ' + bits1 ≠ [] ⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1))+1) + ~ bits2uint(bitsinvert(true + (bits0 + true ' + bits1).rest) +1) = 0, false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
bits0.first ⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1))+1) = bits2uint(bitsinvert(bits0.rest) + (false ' + bitsinvert(bits1))+1), false ∈ bits0

Rule argument

Additional information

Used simplifier rules: bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ bits + boolvar ' + bits0 ≠ [];% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;% ⊦ (bits + false ' + bits0)+1 = bits + (false ' + bits0)+1;% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits + bits0).rest = bits.rest + bits0;