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Proof step 40 from proof abs-split-by-address-h (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, # buf2 = 2 * NODE_HEADER_SIZE + a.alignedsize, # x0 = # lax, abs-node(buf1, nd), abs-node(buf2, a), abs-nodes(buf3 + buf2 + buf, m + # buf1, x0 + a ' + x, lax + @((m + # buf1) + # buf3, # buf2) ' + lax1, boolvar0), ∀ buf4, m, x, lax0, ladr, lax1, boolvar0. # buf4 < #(buf1 + buf3 + buf2 + buf) → abs-nodes(buf4, m, x, lax0 + ladr ' + lax1, boolvar0) → (∃ buf3, buf1, buf5, x0, a, x1. x = x0 + a ' + x1 ∧ # x0 = # lax0 ∧ buf4 = buf3 + buf1 + buf5 ∧ ladr = @(m + # buf3, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf1 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf1, a)) ⊦ ∃ buf6, buf5, buf4, x3, a0, x1. nd ' + x0 + a ' + x = x3 + a0 ' + x1 ∧ # x3 = # lax + 1 ∧ buf1 + buf3 + buf2 + buf = buf6 + buf5 + buf4 ∧ @((m + # buf1) + # buf3, # buf2) = @(m + # buf6, 2 * NODE_HEADER_SIZE + a0.alignedsize) ∧ # buf5 = 2 * NODE_HEADER_SIZE + a0.alignedsize ∧ abs-node(buf5, a0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf1 + buf3, buf2, buf, nd ' + x0, a, x

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(a ' + x) = (# x)+1; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ #(buf + buf0) = # buf + # buf0; (s)% ⊦ @(m, m0) = @(m1, m2) ↔ m = m1 ∧ m0 = m2; (ws)% ⊦ abs-node(buf, nd0) → (abs-node(buf, nd1) ↔ nd0 = nd1); (s) Used terms: buf1 + buf3, buf2, buf, nd ' + x0, a, x