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Proof step 3 from proof is-encoded-app (previous step)

∀ nx1, buf. is-encoded(nx + nx1, buf) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ is-encoded(nx, buf0) ∧ is-encoded(nx1, buf1)) ⊦ (∃ buf0, buf1. buf = buf0 + buf1 ∧ is-encoded(rnode(m), buf0) ∧ is-encoded(nx + nx1, buf1)) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ (∃ buf2, buf1. buf0 = buf2 + buf1 ∧ is-encoded(rnode(m), buf2) ∧ is-encoded(nx, buf1)) ∧ is-encoded(nx1, buf1))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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