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Proof step 21 from proof abs-nodes-elems-adr-length-h (previous step)

# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, ladr = @(m, # buf1), lax = ladr + lax0, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax0, weak), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → # x = # lax ⊦ # x0 + 1 = # lax, lax = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# buf1 = nd.alignedsize + 2 * NODE_HEADER_SIZE, abs-node(buf1, nd), abs-nodes(buf0, m + # buf1, x0, lax0, weak), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → # x = # lax ⊦ # x0 = # lax0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ #(ladr ' + lax) = (# lax)+1; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ ladr ' + lax ≠ []; (s)% ⊦ ladr + lax = ladr ' + lax; (s)