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Proof step 56 from proof eqex-tput (previous step)

st -- r = st0 -- r, st0[rk, val] -- r = st1 -- r, eqex(st0, st, r), eqex(st1, st0[rk, val], r) ⊦ ∀ rk0. (rk0 ∈ st[rk, val] -- r ↔ rk0 ∈ st1 -- r) ∧ (st[rk, val] -- r)[rk0] = (st1 -- r)[rk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
st0[rk, val] -- r = st1 -- r, st -- r = st0 -- r, eqex(st1, st0[rk, val], r), eqex(st0, st, r) ⊦ (rk0 ∈ st[rk, val] -- r ↔ rk0 ∈ st1 -- r) ∧ (st[rk, val] -- r)[rk0] = (st1 -- r)[rk0]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]