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Proof step 24 from proof lastn-fill-sgn (previous step)

¬(true ∈ bits1 ∧ false ∈ bits1), # bits0 = # fillsgn(# bits0, sgn(bits1 + bits0)), # sgn(bits1 + bits0) + 1 < # bits1 + # bits0, # sgn(bits1 + bits0) < # bits0, # sgn(bits1 + bits0) < # bits1 + # bits0 ⊦ bits1 = [], bits1 + bits0 = [], fillsgn(# bits0, sgn(bits1 + bits0)) = bits0, bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
# bits0 = # fillsgn(# bits0, sgn(bits1 + bits0)), # sgn(bits1 + bits0) < # bits1 + # bits0, # sgn(bits1 + bits0) < # bits0, # sgn(bits1 + bits0) + 1 < # bits1 + # bits0 ⊦ bits0 = [], fillsgn(# bits0, sgn(bits1 + bits0)) = bits0, bits1 + bits0 = [], bits1 = [], true ∈ bits1 ∧ false ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;