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Proof step 7 from proof shiftzero-not-neg (previous step)

bits2int(sublist(0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i)))) < 0, i < 0 ⊦ j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ bits ≠ [] → (bits2int(bits) < 0 ↔ bits.first)
Proof step 9:
sublist(0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i))) ≠ [], sublist(0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i))).first, i < 0 ⊦ j = 0

Rule argument

seq: ⊦ bits ≠ [] → (bits2int(bits) < 0 ↔ bits.first) subst: [bits] → [sublist (0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i)))] rotate: F spec: int2bits inst: name: bits2int-lez

Additional information

Used simplifier rules: ⊦ sublist(0, n, bits) = [] ↔ n = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ fillsgn(n, bits) = [] ↔ bits = [] ∧ n = 0;% ⊦ int2bits(j) ≠ []; Used terms: sublist (0, # fillsgn(n, int2bits(i)), mklist(false, i→n(j)) + fillsgn(n, int2bits(i)))