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Proof step 11 from proof abs-nodes-no-padding-last (previous step)

# buf = 2 * NODE_HEADER_SIZE + a.alignedsize, abs-node(buf, a), ∀ buf0, x, a, m, lax, ladr. # buf0 < # buf → # buf0 = (x + a ').flashsize ∧ abs-nodes(buf0, m, x + a ', lax + ladr ', false) → (∃ buf2, buf1. buf0 = buf2 + buf1 ∧ abs-nodes(buf2, m, x, lax, false) ∧ abs-node(buf1, a) ∧ ladr = @(# buf2 + m, 2 * NODE_HEADER_SIZE + a.alignedsize)) ⊦ ∃ buf0, buf1. buf = buf0 + buf1 ∧ abs-nodes(buf0, m, [], [], false) ∧ abs-node(buf1, a) ∧ @(m, # buf0 + # buf1) = @(# buf0 + m, # buf0 + # buf1)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: mkbuf(0), buf abort: F discard: T computed: F

Additional information

Used simplifier rules: abs-nodes-mkbuf0-empty-h: ⊦ abs-nodes(mkbuf(0), m, x, lax, weak) ↔ x = [] ∧ lax = [];% ⊦ mkbuf(0) + buf = buf; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ abs-node(buf, nd0) → (abs-node(buf, nd1) ↔ nd0 = nd1); (s)% ⊦ # mkbuf(n) = n; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: mkbuf(0), buf