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Proof step 10 from proof bitsor-bi-ones (previous step)

1 + # bits2 < # bits ⊦ bits2int(bitsor(bits, true ' + bits2) + bits1) = bits2int(true ' + bits1), bits = [], false ∈ bits

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 11:
n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n
Proof step 12:
∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = # bits2, 1 + # bits2 < # bits ⊦ bits2int(bitsor(bits, true ' + bits2) + bits1) = bits2int(true ' + bits1), bits = [], false ∈ bits

Rule argument

seq: n ≤ # bits ⊦ ∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = n subst: [n, bits] → [# bits2, bits] rotate: F spec: list inst: bitlista name: split-rest

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ m +1 = m + 1;% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: # bits2, bits