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Proof step 4 from proof bits2int-zero-ls (previous step)

¬ 0 ≤ bits2int(bytes2bits(bytes0)) ↔ b2i(bytes0.first) < 0 ⊦ bits2int(byte2bits(by) + bytes2bits(bytes0)) < 0 ↔ ¬ 0 ≤ b2i(by), bytes0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ bits ≠ [] → (bits2int(bits) < 0 ↔ bits.first)
Proof step 6:
byte2bits(by) + bytes2bits(bytes0) ≠ [], ¬ 0 ≤ bits2int(bytes2bits(bytes0)) ↔ b2i(bytes0.first) < 0 ⊦ (byte2bits(by) + bytes2bits(bytes0)).first ↔ ¬ 0 ≤ b2i(by), bytes0 = []

Rule argument

seq: ⊦ bits ≠ [] → (bits2int(bits) < 0 ↔ bits.first) subst: [bits] → [byte2bits(by) + bytes2bits(bytes0)] rotate: F spec: int2bits inst: name: bits2int-lez

Additional information

Used simplifier rules: byte2bits-empty: ⊦ byte2bits(by) ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: byte2bits(by) + bytes2bits(bytes0)