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Proof step 12 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, subarray(buf2, 0, # buf0) = buf0, is-encoded(a, subarray(buf2, 0, # buf0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n])
Proof step 14:
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)), # subarray(buf2, 0, # buf0) = # buf0 ∧ (∀ n. n < # subarray(buf2, 0, # buf0) → subarray(buf2, 0, # buf0)[n] = buf0[n])

Rule argument

spec: array, inst: buffer name: Extension seq: ⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n]) subst: [buf0, buf] → [buf0, subarray(buf2, 0, # buf0)] flags: <all paths>

Additional information

Used terms: buf0, subarray(buf2, 0, # buf0)