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Proof step 15 from proof abs-nodes-case-distinction-h (previous step)

abs-nodes(buf0, m + # buf1, nd ' + x0, @(m0, m1) ' + lax0, true), abs-padding(buf1) ⊦ ∃ buf, buf2. buf1 + buf0 = buf + buf2 ∧ ( buf1 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, @(m0, m1) ' + lax0, true) ∨ (∃ lax1. @(m0, m1) ' + lax0 = @(m, 2 * NODE_HEADER_SIZE + nd.alignedsize) ' + lax1) ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, m + # buf, x0, lax0, true))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf1, buf0

Additional information

Used simplifier rules: abs-nodes-unique-weak-h: ⊦ abs-nodes(buf, m, x0, lax0, true) → (abs-nodes(buf, m, x1, lax1, true) ↔ x0 = x1 ∧ lax0 = lax1); Used terms: buf1, buf0