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Proof step 8 from proof abs-nodes-aligned-headersize-h (previous step)

# buf = 0 ∧ [] = [] ∨ (∃ buf0, buf1. buf = buf0 + buf1 ∧ abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, [], [], weak)) ∨ weak ∧ abs-failed(buf) ∧ [] = [], ∀ buf0, m, x, lax, weak. # buf0 < # buf → abs-nodes(buf0, m, x, lax, weak) → is-aligned(# buf0, NODE_HEADER_SIZE) ⊦ is-aligned(# buf, NODE_HEADER_SIZE)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
buf = mkbuf(0) ∨ buf = buf0 + buf1 ∧ abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, [], [], weak) ∨ weak ∧ abs-failed(buf), ∀ buf0, m, x, lax, weak. # buf0 < # buf → abs-nodes(buf0, m, x, lax, weak) → is-aligned(# buf0, NODE_HEADER_SIZE) ⊦ is-aligned(# buf, NODE_HEADER_SIZE)

Rule argument

Additional information

Used simplifier rules: ⊦ # buf = 0 ↔ buf = mkbuf(0); (s)