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Proof step 4 from proof fill-nls (
previous step
)
¬(# bits ≤ n ∧ bits ≠ []), # fillsgn(n, bits) < # bits ⊦ bits = []
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
# fillsgn(n, bits) < # bits, n < # bits ⊦ bits = []
Rule argument
Additional information
Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;