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Proof step 157 from proof byte-law-and-one (previous step)

int2bits(j) = bits0 + bits, # bits = 8, 7 = # fillsgn(7, int2bits(i)), # int2bits(i) < # bitsand(bits, fillsgn(8, int2bits(i))), i < 0, # int2bits(i) < 8, 8 ≤ # bits0 + # bitsand(bits, fillsgn(8, int2bits(i))), # int2bits(i) < # bits0 + # bitsand(bits, fillsgn(8, int2bits(i))), 8 ≤ # bitsand(bits, fillsgn(8, int2bits(i))) ⊦ bits0 + bits = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(i)))) = lastn(8, bitsand(bits, true ' + fillsgn(7, int2bits(i)))), bitsand(bits, true ' + fillsgn(7, int2bits(i))) = []

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

seq: n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n subst: [n, bits] → [# int2bits(i), bits] rotate: F spec: list inst: bitlista name: split-rest

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) Used terms: # int2bits(i), bits