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Proof step 9 from proof eqex-trans (conclusion)

⊦ eqex(st, st0, r) ∧ eqex(st0, st1, r) → eqex(st, st1, r)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ eqex(st, st0, r) → eqex(st0, st, r)
Proof step 8:
eqex(st1, st0, r), eqex(st0, st, r), eqex(st0, st1, r), eqex(st, st0, r) ⊦ eqex(st, st1, r)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]