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Proof step 6 from proof newrefs-empty (previous step)

∀ st. newref_list(m, st) = [] ↔ m = 0 ⊦ newref_list(m +1, st) = [] ↔ m +1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
newref_list(m + 1, st) = [], ∀ st. newref_list(m, st) = [] ↔ m = 0 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ m +1 = m + 1])]]