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Proof step 6 from proof bits2uint-not-neg (previous step)

∀ bits0. # bits0 < # bits → true → (bits2uint(bits0) ≤ 0 ↔ bits2uint(bits0) = 0) ⊦ ∀ bits0, boolvar. bits ≠ bits0 + boolvar ', bits2uint(bits) ≤ 0 ↔ bits2uint(bits) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits1. # bits1 < #(bits0 + boolvar ') → true → (bits2uint(bits1) ≤ 0 ↔ bits2uint(bits1) = 0) ⊦ bits2uint(bits0 + boolvar ') ≤ 0 ↔ bits2uint(bits0 + boolvar ') = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;