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Proof step 4 from proof indel (conclusion)

⊦ r ∈ st ∧ r ≠ r0 → r ∈ st -- (r0 -- sk)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r ∈ st ↔ (∃ sk. r -- sk ∈ st)
Proof step 2:
⊦ r -- sk ∈ st → r ∈ st
Proof step 3:
r -- sk0 ∈ st, r -- sk1 ∈ st ⊦ r = r0, ∃ sk0. r -- sk0 ∈ st -- (r0 -- sk)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]