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Proof step 28 from proof abs-split-by-address-h (previous step)

# buf2 = nd.alignedsize + 2 * NODE_HEADER_SIZE, abs-node(buf2, nd), abs-nodes(buf, (m + # buf1) + # buf2, x2, lax3, boolvar0), abs-padding(buf1), ∀ buf3, m, x, lax0, ladr, lax1, boolvar0. # buf3 < #(buf1 + buf2 + buf) → abs-nodes(buf3, m, x, lax0 + ladr ' + lax1, boolvar0) → (∃ buf5, buf1, buf4, x0, a, x1. x = x0 + a ' + x1 ∧ # x0 = # lax0 ∧ buf3 = buf5 + buf1 + buf4 ∧ ladr = @(m + # buf5, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf1 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf1, a)) ⊦ ∃ buf5, buf4, buf3, a, x0. nd ' + x2 = a ' + x0 ∧ buf1 + buf2 + buf = buf5 + buf4 + buf3 ∧ @(m + # buf1, # buf2) = @(m + # buf5, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf4 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf4, a)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf1, buf2, buf, nd, x2

Additional information

Used simplifier rules: append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ @(m, m0) = @(m1, m2) ↔ m = m1 ∧ m0 = m2; (ws)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ abs-node(buf, nd0) → (abs-node(buf, nd1) ↔ nd0 = nd1); (s) Used terms: buf1, buf2, buf, nd, x2