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Proof step 3 from proof abs-nodes-split (previous step)

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, # buf0, x1, lax1, weak)

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ abs-nodes(buf, m, x0 + x1, lax, weak) → (∃ buf0, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf = buf0 + buf1 ∧ abs-nodes(buf0, m, x0, lax0, false) ∧ abs-nodes(buf1, m + # buf0, x1, lax1, weak))
Proof step 5:
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)

Rule argument

spec: <toplevel> name: abs-nodes-split-h seq: ⊦ abs-nodes(buf, m, x0 + x1, lax, weak) → (∃ buf0, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf = buf0 + buf1 ∧ abs-nodes(buf0, m, x0, lax0, false) ∧ abs-nodes(buf1, m + # buf0, x1, lax1, weak)) subst: [buf, m, x0, x1, lax, weak] → [buf, 0, x0, x1, lax, weak]

Additional information

Used terms: buf, 0, x0, x1, lax, weak