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Proof step 38 from proof lastn-fill-sgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 39:
⊦ bits ≠ [] ∧ (bits.first ↔ ¬ boolvar) → sgn(boolvar ' + bits) = boolvar ' + bits
Proof step 40:
¬(bits0 ≠ [] ∧ (bits0.first ↔ ¬ false)), # bits0 = # fillsgn(# bits0, sgn(false ' + bits0)), # sgn(false ' + bits0) + 1 < # bits1 + # bits0, # sgn(false ' + bits0) < # bits0, # sgn(false ' + bits0) < # bits1 + # bits0 ⊦ bits1 = [], bits1 + bits0 = [], fillsgn(# bits0, sgn(false ' + bits0)) = bits0, bits0 = [], true ∈ bits1

Rule argument

name: sgn-base-03 rotate: F subst: [bits, boolvar] → [bits0, false]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ ¬ m + n < m; Used terms: bits0, false