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Proof step 22 from proof lastn-fill-sgn (previous step)

# 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 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ true ∈ bits ∧ false ∈ bits → sgn(bits + bits0) = sgn(bits) + bits0
Proof step 24:
¬(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 = []

Rule argument

name: sgn-append-01 rotate: F subst: [bits0, bits] → [bits0, bits1]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (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, bits1