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Proof step 11 from proof is-encoded-prefix (previous step)

buf0 + buf1 = buf2 + buf3, is-encoded(a, buf0), # buf0 ≤ # buf2 ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b, is-encoded(a, subarray(buf2, 0, # buf0))

User applied rule cut formula

And obtained the following new goals

Proof step 12:
buf0 + buf1 = buf2 + buf3, is-encoded(a, buf0), # buf0 ≤ # buf2 ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b, subarray(buf2, 0, # buf0) = buf0, is-encoded(a, subarray(buf2, 0, # buf0))

Rule argument

fma: subarray(buf2, 0, # buf0) = buf0

Additional information

Used formulas: subarray(buf2, 0, # buf0) = buf0