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Proof step 5 from proof decode (previous step)

n0 = 0, is-encoded(vtbl0, buf) ⊦ ⟪decode_nat#(n0, buf; n1, n; err); n0 := n0 + n; vtbl := ∅; 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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
is-encoded(vtbl0, buf) ⊦ ⟪decode_nat#(0, buf; n1, n; err); n0 := n; vtbl := ∅; 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))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)