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Proof step 5 from proof abs-nodes-split (previous step)

abs-nodes(buf, 0, x0 + x1, lax, weak), abs-nodes(buf, 0, x0 + x1, lax, weak) → (∃ buf0, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf = buf0 + buf1 ∧ abs-nodes(buf0, 0, x0, lax0, false) ∧ abs-nodes(buf1, 0 + # buf0, x1, lax1, weak)) ⊦ ∃ buf0, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf = buf0 + buf1 ∧ abs-nodes(buf0, 0, x0, lax0, false) ∧ abs-nodes(buf1, # buf0, x1, lax1, weak)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
abs-nodes(buf1, # buf0, x1, lax1, weak), abs-nodes(buf0, 0, x0, lax0, false), abs-nodes(buf0 + buf1, 0, x0 + x1, lax0 + lax1, weak) ⊦ ∃ buf2, buf3, lax2, lax3. lax0 + lax1 = lax2 + lax3 ∧ buf0 + buf1 = buf2 + buf3 ∧ abs-nodes(buf2, 0, x0, lax2, false) ∧ abs-nodes(buf3, # buf2, x1, lax3, weak)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)