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Proof step 23 from proof abs-nodes-nospace-h (previous step)

lax.head = @(m0, m1), # buf1 + # buf0 < NODE_HEADER_SIZE, buf1 = buf3 + buf4 + validtrailer ∧ is-encoded(nodeheader(# buf4, true), buf3) ∧ is-aligned(# buf4, 2 * NODE_HEADER_SIZE) ∧ abs-nodes(buf0, m + # buf3 + # buf4 + NODE_HEADER_SIZE, nd ' + x0, lax, boolvar) ∨ lax ≠ [] ∧ m0 = m ∧ m1 = # buf + # buf2 + NODE_HEADER_SIZE ∧ buf1 = buf + buf2 + validtrailer ∧ is-encoded(nodeheader(node-size-headerless(nd), false), buf) ∧ is-encoded(nd, buf2) ∧ # buf2 = nd.alignedsize ∧ abs-nodes(buf0, m + # buf + # buf2 + NODE_HEADER_SIZE, x0, lax.tail, boolvar) ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: abs-nodes-first-adr-explicit: ⊦ abs-nodes(buf, m, a ' + x, lax, weak) → (lax.head = ladr ↔ (∃ lax0. lax = ladr + lax0));% ⊦ #(buf + buf0) = # buf + # buf0; (s)% ⊦ # validtrailer = NODE_HEADER_SIZE; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ ladr + lax = ladr ' + lax; (s)