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Proof step 10 from proof abs-nodes-no-padding-one (previous step)

# buf0 = a.alignedsize + 2 * NODE_HEADER_SIZE, ladr.size = # buf0, lax = ladr + lax0, lax0 = [], abs-node(buf0, a) ⊦ lax = @(ladr.off, # buf0) ', lax = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# buf0 = a.alignedsize + 2 * NODE_HEADER_SIZE, ladr.size = # buf0, abs-node(buf0, a) ⊦ ladr = @(ladr.off, # buf0)

Rule argument

Additional information

Used simplifier rules: ⊦ ladr ' = ladr0 ' ↔ ladr = ladr0; (s)% ⊦ ladr ' ≠ []; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ lax + [] = lax; (s)% ⊦ ladr + lax = ladr ' + lax; (s)