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Proof step 20 from proof is-encoded-prefix (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 21:
⊦ is-encoded(a, buf) ∧ n ≤ # buf → (is-encoded(b, subarray(buf, 0, n)) ↔ b = a ∧ n = # buf)
Proof step 22:
buf0 + buf1 = buf2 + buf3, # buf2 < # buf0, is-encoded(a, buf0), is-encoded(a, buf0) ∧ # buf2 ≤ # buf0 → (is-encoded(b, subarray(buf0, 0, # buf2)) ↔ b = a ∧ # buf2 = # buf0) ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b

Rule argument

spec: <toplevel> name: is-encoded-unique-size seq: ⊦ is-encoded(a, buf) ∧ n ≤ # buf → (is-encoded(b, subarray(buf, 0, n)) ↔ b = a ∧ n = # buf) subst: [a, buf, n, b] → [a, buf0, # buf2, b]

Additional information

Used terms: a, buf0, # buf2, b