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Proof step 4 from proof abs-nodes-one-h (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a + x = a ' + x
Proof step 6:
⊦ abs-nodes(buf, m, a ' + [], lax, weak) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, a ' + [], lax, weak) ∨ lax = @(m, 2 * NODE_HEADER_SIZE + a.alignedsize) ' ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, [], [], weak)))

Rule argument

spec: list name: c seq: ⊦ a + x = a ' + x subst: [x, a] → [[], a] flags: <all paths>

Additional information

Used terms: [], a