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Proof step 25 from proof shiftzero-zero-size (previous step)

i < 0, firstn(# int2bits(i), mklist(false, i→n(j)) + int2bits(i)).first ⊦ j = 0, bits2int(firstn(# int2bits(i), mklist(false, i→n(j)) + int2bits(i))) = bits2uint(firstn(# int2bits(i), mklist(false, i→n(j)) + int2bits(i)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ n ≠ 0 ∧ bits ≠ [] → (firstn(n, bits).first ↔ bits.first)
Proof step 27:
# int2bits(i) ≠ 0 ∧ mklist(false, i→n(j)) + int2bits(i) ≠ [], i < 0, (mklist(false, i→n(j)) + int2bits(i)).first ⊦ j = 0, bits2int(firstn(# int2bits(i), mklist(false, i→n(j)) + int2bits(i))) = bits2uint(firstn(# int2bits(i), mklist(false, i→n(j)) + int2bits(i)))

Rule argument

seq: ⊦ n ≠ 0 ∧ bits ≠ [] → (firstn(n, bits).first ↔ bits.first) subst: [bits, n] → [mklist(false, i→n(j)) + int2bits(i), # int2bits(i)] rotate: F spec: list-del inst: bitlista name: firstn-first

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: mklist(false, i→n(j)) + int2bits(i), # int2bits(i)