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Proof step 9 from proof bits2uint-succ (previous step)

boolvar, ∀ bits1. # bits1 < #(bits0 + boolvar ') → true → bits1 = [] ∨ bits2uint(bits1 +1) = bits2uint(bits1) + 1 ⊦ bits2uint(bits0 +1 + false) = bits2uint(bits0 + boolvar ') + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ bits1. # bits1 < #(bits0 + true ') → true → bits1 = [] ∨ bits2uint(bits1 +1) = bits2uint(bits1) + 1 ⊦ bits2uint(bits0 +1) = bits2uint(bits0) + 1

Rule argument

Additional information

Used simplifier rules: bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;%bits2uint-one: ⊦ bits2uint(bits + true ') = bits2uint(bits) * 2 + 1;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i = i * (j + 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar = bits + boolvar ';