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Proof step 28 from proof abs-nodes-append-h (previous step)

abs-nodes(buf + buf1, m + # buf2, (nd ' + x) + x1, (@(m2, m3) ' + lax) + lax1, weak), abs-nodes(buf1, (m + # buf2) + # buf, x1, lax1, weak), abs-nodes(buf, m + # buf2, nd ' + x, @(m2, m3) ' + lax, false), abs-padding(buf2), ∀ buf3, m, x0, lax0, buf1, x1, lax1, weak. # buf3 < #(buf2 + buf) → abs-nodes(buf3, m, x0, lax0, false) → (abs-nodes(buf3 + buf1, m, x0 + x1, lax0 + lax1, weak) ↔ abs-nodes(buf1, m + # buf3, x1, lax1, weak)) ⊦ ∃ buf0, buf3. (buf2 + buf) + buf1 = buf0 + buf3 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf3, m + # buf0, nd ' + x + x1, (@(m2, m3) ' + lax) + lax1, weak) ∨ m2 = m ∧ m3 = # buf0 ∧ abs-node(buf0, nd) ∧ abs-nodes(buf3, m + # buf0, x + x1, lax + lax1, weak))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf2, buf + buf1

Additional information

Used simplifier rules: append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]%a: ⊦ (x + y) + z = x + y + z; from specification list%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list] Used terms: buf2, buf + buf1