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Proof step 28 from proof is-encoded-prefix (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 29:
buf0 + buf1 = buf2 + buf3, # buf2 < # buf0, is-encoded(b, buf2), is-encoded(a, buf0) ⊦ # buf0 = # buf2, subarray(buf0, 0, # buf2) = buf2, is-encoded(b, subarray(buf0, 0, # buf2))

Rule argument

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

Additional information

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