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Proof step 2 from proof abs-nodes-crash (previous step)

abs-nodes(buf, x, lax, true) ⊦ ∃ x0, lax0. x0 ⊑ x ∧ lax0 ⊑ lax ∧ abs-nodes(buf.align↓, x0, lax0, true)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
abs-nodes(buf, 0, x, lax, true) ⊦ ∃ x0, lax0. x0 ⊑ x ∧ lax0 ⊑ lax ∧ abs-nodes(buf.align↓, 0, x0, lax0, true)

Rule argument

Additional information

Used simplifier rules: abs-nodes: ⊦ abs-nodes(buf, x, lax, weak) ↔ abs-nodes(buf, 0, x, lax, weak);