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Proof step 6 from proof succ-zeros-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits2, bits. # bits2 < #(bits1 + true ') → true → (bits + bits2)+1 = bits +1 + bitsinvert(bits2) ∨ false ∈ bits2 ⊦ (bits + bits1)+1 = bits +1 + bitsinvert(bits1), false ∈ bits1

Rule argument

Additional information

Used simplifier rules: invert-base: ⊦ bitsinvert([]) = [];%inv-last: ⊦ bitsinvert(bits + boolvar ') = bitsinvert(bits) + (¬ boolvar)';%succ-rec-01: ⊦ boolvar → (bits + boolvar ')+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)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;