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Proof step 5 from proof succ-zero-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ bits2, bits. # bits2 < #(bits1 + boolvar ') → false ∈ bits2 → (bits + bits2)+1 = bits + bits2 +1 ⊦ (bits + bits1 + boolvar ')+1 = bits + (bits1 + boolvar ')+1, boolvar ∧ ¬ false ∈ bits1

Rule argument

Additional information

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