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Proof step 34 from proof bits2int-lasteq (previous step)

true ∧ bits0.first ∧ bits0 ≠ [] ⊦ bits0 = [], ~1 = 2 * bits2uint(bitsinvert(bits1)) + ~(bits2uint(bitsinvert(bits0.rest)) * 2 + 1) ↔ ~1 = bits2int(bits0) + bits2uint(bitsinvert(bits1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
bits0.first ⊦ bits0 = [], bits2uint(bitsinvert(bits1)) = bits2uint(bitsinvert(bits0.rest)) ↔ ~1 = bits2int(bits0) + bits2uint(bitsinvert(bits1))

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;