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Proof step 4 from proof succ-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ bits2, bits. # bits2 < #(bits1 + boolvar ') → true → (bits + false ' + bits2)+1 = bits + (false ' + bits2)+1 ⊦ (bits + false ' + bits1 + boolvar ')+1 = bits + (false ' + bits1 + boolvar ')+1

Rule argument

Additional information

Used simplifier rules: succ-zero: ⊦ (bits + false ')+1 = bits + true;%succ-one: ⊦ false ' +1 = true ';% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar ' ≠ [];% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;