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Proof step 21 from proof ADD-lemma (previous step)

s0 ≠ ∅, s0 ⊆ S, S3 = V0 ∪ (S \ s0), S2 = N0 ∪ (S \ (s0 ∪ V0)), s0 = s, w ∈ s0, ¬ w ∈ S3, ¬ s0 -- w ⊆ S ∨ S3 ++ w ≠ V0 ∪ (S \ (s0 -- w)) ∨ S2 ++ w ≠ N0 ∪ (S \ ((s0 -- w) ∪ V0)) ∨ ¬ # (s0 -- w) < # s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
s ⊆ S, w ∈ s, w ∈ S, (V0 ∪ (S \ s)) ++ w ≠ V0 ∪ (S \ (s -- w)) ∨ (N0 ∪ (S \ (s ∪ V0))) ++ w ≠ N0 ∪ (S \ ((s -- w) ∪ V0)) ⊦ S = ∅, s = ∅, # s = 0, # S = 0, w ∈ V0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ ¬ n < n], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0], [CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : ⊦ m < 1 ↔ m = 0], [CSIMPFORWARD : ⊦ # s = 0 → s = ∅], [CSIMPFORWARD : ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0], [CSIMPFORWARD : ⊦ a ∈ s → s ≠ ∅], [CSIMPFORWARD : ⊦ s ≠ ∅ → # s ≠ 0], [CSIMPSEQ : ⊦ a ∈ s → # (s -- a) = # s - 1], [CSIMPSEQ : ⊦ s ⊆ s0 → s -- a ⊆ s0], [CSIMPSEQ : ⊦ s ∪ s0 = s0 ∪ s], [CSIMPSEQ : ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1], [CSIMPSEQ : ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2])]]