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Proof step 6 from proof bits2uint-succ (previous step)

∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ bits2uint(bits0 +1) = bits2uint(bits0) + 1 ⊦ bits2uint(bits +1) = bits2uint(bits) + 1, ∀ bits0, boolvar. bits ≠ bits0 + boolvar '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits1. # bits1 < #(bits0 + boolvar ') → true → bits1 = [] ∨ bits2uint(bits1 +1) = bits2uint(bits1) + 1 ⊦ bits2uint((bits0 + boolvar ')+1) = bits2uint(bits0 + boolvar ') + 1

Proof step 6 from proof bits2uint-succ (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information