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Proof step 9 from proof fillsgn-short (previous step)

2 < # fillsgn(# bytes, int2bytes(bytes2int(bytes))), # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))), 2 < # bytes ⊦ fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ int2bytes(bytes2int(bytes)) = bytes, bytes = [], fillsgn(# bytes, int2bytes(bytes2int(bytes))) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ # bytes ≥ n → fillsgn(n, bytes) = bytes
Proof step 11:
¬ # int2bytes(bytes2int(bytes)) ≥ 2, 2 < # fillsgn(# bytes, int2bytes(bytes2int(bytes))), # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))), 2 < # bytes ⊦ fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ int2bytes(bytes2int(bytes)) = bytes, bytes = [], fillsgn(# bytes, int2bytes(bytes2int(bytes))) = []

Rule argument

name: fillsgn-longer rotate: F subst: [bytes, n] → [int2bytes(bytes2int(bytes)), 2]

Additional information

Used terms: int2bytes(bytes2int(bytes)), 2