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Proof step 6 from proof is-encoded-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ is-encoded(a, buf) → (is-encoded(b, buf) ↔ a = b)
Proof step 8:
buf0 + buf1 = buf2 + buf3, is-encoded(a, buf0), # buf0 ≤ # buf2, is-encoded(b, buf2) → is-encoded(a, subarray(buf2, 0, # buf0)) → a = b ∧ # buf0 = # buf2 ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b

Rule argument

Additional information

Used simplifier rules: is-encoded-unique: ⊦ is-encoded(a, buf) → (is-encoded(b, buf) ↔ a = b);%# buf = n ⊦ subarray(buf, 0, n) = buf; (s)