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Proof step 6 from proof abs-nodes-split-h (previous step)

abs-nodes(buf, m, x1, lax, weak), ∀ buf0, m, x0, x1, lax, weak. # buf0 < # buf → abs-nodes(buf0, m, x0 + x1, lax, weak) → (∃ buf2, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf0 = buf2 + buf1 ∧ abs-nodes(buf2, m, x0, lax0, false) ∧ abs-nodes(buf1, m + # buf2, x1, lax1, weak)) ⊦ ∃ buf0, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf = buf0 + buf1 ∧ abs-nodes(buf0, m, [], lax0, false) ∧ abs-nodes(buf1, m + # buf0, x1, lax1, weak)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: mkbuf(0), buf, [], lax 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 = [];% ⊦ [] + lax = lax; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ mkbuf(0) + buf = 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 Used terms: mkbuf(0), buf, [], lax