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Proof step 35 from proof abs-nodes-subarray-failed (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, # buf0 = x0.flashsize, abs-node(buf1, nd), abs-nodes(buf0, m0 + # buf1, x0, lax0, false), ∀ buf2, x, n, m, lax. # buf2 < #(buf1 + buf0) → # buf2 = x.flashsize ∧ n < x.flashsize ∧ abs-nodes(buf2, m, x, lax, false) ∧ is-aligned(n, NODE_HEADER_SIZE) → n = 0 ∨ x = [] ∨ (∃ n0. n0 < # lax ∧ abs-nodes(subarray(buf2, 0, n), m, prefixn(n0, x), prefixn(n0, lax), true)) ⊦ buf1 + buf0 = mkbuf(0), buf1 = mkbuf(0), lax0 = [], x0 = [], ∃ buf, buf2. buf1 = buf + buf2 ∧ ( abs-padding(buf) ∧ abs-nodes(buf2, m0 + # buf, nd ', @(m0, # buf + # buf2) ', true) ∨ abs-node(buf, nd) ∧ abs-nodes(buf2, m0 + # buf, [], [], true))

User applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: buf1, mkbuf(0) 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 = [];% ⊦ buf + mkbuf(0) = buf; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ # 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% ⊦ a ' ≠ []; (s)% ⊦ abs-node(buf, nd0) → (abs-node(buf, nd1) ↔ nd0 = nd1); (s) Used terms: buf1, mkbuf(0)