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

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

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: boolvar

Additional information

Used simplifier rules: succ-zero: ⊦ (bits + false ')+1 = bits + true;%bool freely generated by true, false;;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;