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Proof step 7 from proof abs-nodes-emptyheader-h (previous step)

isempty(buf, 0, NODE_HEADER_SIZE) ⊦ # buf = 0 ∧ lax = [] ∨ (∃ buf0, buf1. buf = buf0 + buf1 ∧ abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, [], lax, weak)) ∨ weak ∧ abs-failed(buf) ∧ lax = [] ↔ (buf = mkbuf(0) ∨ abs-failed(buf) ∧ weak) ∧ lax = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
isempty(buf, 0, NODE_HEADER_SIZE) ⊦ buf = mkbuf(0) ∧ lax = [] ∨ (∃ buf0, buf1. buf = buf0 + buf1 ∧ abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, [], lax, weak)) ∨ weak ∧ abs-failed(buf) ∧ lax = [] ↔ (buf = mkbuf(0) ∨ abs-failed(buf) ∧ weak) ∧ lax = []

Rule argument

Additional information

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