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Proof step 3 from proof bits2uint-not-neg (previous step)

∀ bits0. # bits0 < # bits → true → (bits2uint(bits0) ≤ 0 ↔ bits2uint(bits0) = 0) ⊦ bits2uint(bits) ≤ 0 ↔ bits2uint(bits) = 0

User applied rule cut formula

And obtained the following new goals

Proof step 4:
∀ bits0. # bits0 < # bits → true → (bits2uint(bits0) ≤ 0 ↔ bits2uint(bits0) = 0) ⊦ bits = [], bits2uint(bits) ≤ 0 ↔ bits2uint(bits) = 0

Rule argument

fma: bits = []

Additional information

Used simplifier rules: bits2uint-base: ⊦ bits2uint([]) = 0; Used formulas: bits = []