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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
abs-failed(buf), buf = mkbuf(0) ∧ lax = [] ∨ buf = buf0 + buf1 ∧ abs-padding(buf0) ∧ abs-nodes(buf1, n + # buf0, [], lax, weak) ∨ weak ∧ lax = [] ⊦ lax = [] ∧ weak

Rule argument

Additional information

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