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Proof step 5 from proof fillsgn-short (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 6:
¬ # bytes < 2, # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))) ⊦ bytes = [], fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ ¬ # bytes < 2 ∧ (2 < # bytes → int2bytes(bytes2int(bytes)) = bytes)
Proof step 20:
# bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))) ⊦ ¬ # bytes < 2, bytes = [], fillsgn(2, int2bytes(bytes2int(bytes))) = bytes ↔ ¬ # bytes < 2 ∧ (2 < # bytes → int2bytes(bytes2int(bytes)) = bytes)

Rule argument

fma: ¬ # bytes < 2

Additional information

Used formulas: ¬ # bytes < 2