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Proof step 14 from proof intshiftright-ls-01 (previous step)

j < 32, 0 ≤ j ⊦ bits2int((mklist(i < 0, i→n(j)) + int2bits(i)).first ' + sublist(0, # int2bits(i) - 1, (mklist(i < 0, i→n(j)) + int2bits(i)).rest)) < 0 ↔ ¬ 0 ≤ i, mklist(¬ 0 ≤ i, i→n(j)) + int2bits(i) = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
# int2bits(i) = m + 1 ∧ # int2bits(i) ≠ 0, j < 32, 0 ≤ j ⊦ bits2int((mklist(i < 0, i→n(j)) + int2bits(i)).first ' + sublist(0, m, (mklist(i < 0, i→n(j)) + int2bits(i)).rest)) < 0 ↔ ¬ 0 ≤ i, mklist(¬ 0 ≤ i, i→n(j)) + int2bits(i) = []

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# int2bits(i), m] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ [];%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # int2bits(i), m