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Proof step 1 from proof succ-exinv (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ boolvar ∈ bits → bits ≠ bits0 + boolvar ' + bits1;