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Proof step 1 from proof abs-nodes-one (conclusion)

⊦ abs-nodes(buf, a ', lax, weak) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, # buf0, a ', lax, weak) ∨ lax = @(0, a.flashsize) ' ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, # buf0, [], [], weak)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ abs-nodes(buf, 0, a ', lax, weak) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, # buf0, a ', lax, weak) ∨ lax = @(0, 2 * NODE_HEADER_SIZE + a.alignedsize) ' ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, # buf0, [], [], weak)))

Rule argument

Additional information

Used simplifier rules: abs-nodes: ⊦ abs-nodes(buf, x, lax, weak) ↔ abs-nodes(buf, 0, x, lax, weak);% ⊦ nd.flashsize = 2 * NODE_HEADER_SIZE + nd.alignedsize; (s)