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Proof step 9 from proof remove-refsize0-one (previous step)

(∅)[n, mkblock(x, lax, m0)][n0] = mkblock(x0, lax0, m) ⊦ (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ m ≠ 0 ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ m ≠ 0 → mkblock(x0, lax0, m) = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 10:
(∅)[n, mkblock(x, lax, m0)][n0] = mkblock(x0, lax0, m) ⊦ n0 = n, (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ m ≠ 0 ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ (n0 ∈ (∅)[n, mkblock(x, lax, m0)] ∧ m ≠ 0 → mkblock(x0, lax0, m) = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Rule argument

fma: n0 = n

Additional information

Used simplifier rules: ⊦ st[n, ablock][n] = ablock; (s)%⊦ mkblock(x, lax, m) = mkblock(x0, lax0, m0) ↔ x = x0 ∧ lax = lax0 ∧ m = m0; (ws)% ⊦ n ∈ st[n, ablock]; (s)% ⊦ ¬ n ∈ ∅; (s) Used formulas: n0 = n