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Proof step 6 from proof sgn-not-empty (
previous step
)
¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), sgn(boolvar ' + bits0) = [] ⊦ bits0 = [], sgn(bits0) = []
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 7
:
sgn(boolvar ' + bits0) = [] ⊦ bits0 = [], sgn(bits0) = [], bits0.first ↔ boolvar
Rule argument
Additional information