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Proof step 2 from proof decode (previous step)

is-encoded(vtbl0, buf) ⊦ ⟪vtbl := ∅; err := ESUCCESS; let n1 = ?, n0 = 0, n = ? in {decode_nat#(n0, buf; n1, n; err); n0 := n0 + n; while n1 ≠ 0 ∧ err = ESUCCESS do {/* (decode : invariant (err = ESUCCESS ∨ ⌊err⌋ ) ∧ ( err = ESUCCESS → (∃ vtbl1. vtbl ⊥ vtbl1 ∧ # vtbl1 + # vtbl = n1` ∧ # vtbl1 = n1 ∧ # buf ≥ ENCODED_NAT_SIZE + n1` * (ENCODED_VOLID_SIZE + ENCODED_NAT_SIZE) ∧ n0 = ENCODED_NAT_SIZE + # vtbl * (ENCODED_VOLID_SIZE + ENCODED_NAT_SIZE) ∧ is-encoded(n1`, subarray(buf, 0, ENCODED_NAT_SIZE)) ∧ is-encoded-rec(vtbl, subarray(buf, ENCODED_NAT_SIZE, # vtbl * (ENCODED_VOLID_SIZE + ENCODED_NAT_SIZE))) ∧ is-encoded-rec(vtbl1, subarray(buf, n0, # vtbl1 * (ENCODED_VOLID_SIZE + ENCODED_NAT_SIZE))))) wfbound n1) */; let v = ?, m = ? in {decode_volid#(n0, buf; v, n; err); if err = ESUCCESS then {n0 := n0 + n; decode_nat#(n0, buf; m, n; err); if err = ESUCCESS then {n0 := n0 + n; vtbl[v] := m}}}; n1 := n1 - 1}}⟫ ((err = ESUCCESS ∨ ⌊err⌋ ) ∧ (err = ESUCCESS → vtbl = vtbl0))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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