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Proof step 19 from proof is-encoded-prefix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
buf0 + buf1 = buf2 + buf3, # buf2 < # buf0, is-encoded(a, buf0) ⊦ is-encoded(b, buf2) ↔ # buf0 = # buf2 ∧ a = b

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)