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Proof step 6 from proof eqex-trans (previous step)

eqex(st1, st0, r), eqex(st0, st, r), eqex(st0, st1, r), st -- r = st0 -- r ⊦ st -- r = st1 -- r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
st -- r = st0 -- r, eqex(st1, st0, r), eqex(st0, st, r), eqex(st0, st1, r) ⊦ st0 -- r = st1 -- r

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]