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Proof step 1 from proof is-encoded-empty (conclusion)

⊦ is-encoded(nx, mkbuf(0)) ↔ nx = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
is-encoded(nx, mkbuf(0)) ⊦ nx = []

Rule argument

Additional information

Used simplifier rules: is-encoded-base: ⊦ is-encoded([], buf) ↔ buf = mkbuf(0);