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Proof step 4 from proof succ-ex (previous step)

∃ bits0, bits1. bits = bits0 + false ' + bits1 ∧ ¬ false ∈ bits1 ⊦ ∃ 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 5:
⊦ false ∈ bits0, ∃ bits, bits2. bits1 + false ' + bits0 = bits + false ' + bits2 ∧ ¬ false ∈ bits2 ∧ bits1 + true ' + bitsinvert(bits0) = bits + true ' + bitsinvert(bits2)

Rule argument

Additional information

Used simplifier rules: succ-prefix: ⊦ (bits + false ' + bits0)+1 = bits + (false ' + bits0)+1;%succ-zeros-01: ⊦ ¬ false ∈ bits0 → (bits + bits0)+1 = bits +1 + bitsinvert(bits0);%succ-one: ⊦ false ' +1 = true ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;