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Proof step 3 from proof is-encoded-prefix (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 4:
buf0 + buf1 = buf2 + buf3, is-encoded(a, buf0), # buf0 ≤ # buf2 ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b
Proof step 19:
buf0 + buf1 = buf2 + buf3, is-encoded(a, buf0) ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b, # buf0 ≤ # buf2

Rule argument

fma: # buf0 ≤ # buf2

Additional information

Used formulas: # buf0 ≤ # buf2