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Proof step 1 from proof succ-ex (conclusion)

⊦ false ∈ bits ↔ (∃ bits0, bits1. bits +1 = bits0 + true ' + bitsinvert(bits1) ∧ bits = bits0 + false ' + bits1 ∧ ¬ false ∈ bits1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
false ∈ bits ⊦ ∃ bits0, bits1. bits +1 = bits0 + true ' + bitsinvert(bits1) ∧ bits = bits0 + false ' + bits1 ∧ ¬ false ∈ bits1

Rule argument

Additional information

Used simplifier rules: succ-ones: ⊦ ¬ false ∈ bits → bits +1 = true + bitsinvert(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ boolvar ∈ bits → bits ≠ bits0 + boolvar ' + bits1;% ⊦ boolvar + bits = boolvar ' + bits;