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Proof step 23 from proof eqex-tput (previous step)

¬ r1 -- sk0 ∈ st[r -- sk, val] -- r1, eqex(st, st0, r1), eqex(st0, st, r1), eqex(st1, st0[r -- sk, val], r1), eqex(st0[r -- sk, val], st1, r1), ∀ rk0. (rk0 ∈ st0[r -- sk, val] -- r1 ↔ rk0 ∈ st1 -- r1) ∧ (st0[r -- sk, val] -- r1)[rk0] = (st1 -- r1)[rk0], ∀ rk. (rk ∈ st -- r1 ↔ rk ∈ st0 -- r1) ∧ (st -- r1)[rk] = (st0 -- r1)[rk] ⊦ ∅[r1 -- sk0] = (st1 -- r1)[r1 -- sk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
eqex(st0[r -- sk, val], st1, r1), eqex(st1, st0[r -- sk, val], r1), eqex(st0, st, r1), eqex(st, st0, r1), ∀ rk. (rk ∈ st -- r1 ↔ rk ∈ st0 -- r1) ∧ (st -- r1)[rk] = (st0 -- r1)[rk], ∀ rk0. (rk0 ∈ st0[r -- sk, val] -- r1 ↔ rk0 ∈ st1 -- r1) ∧ (st0[r -- sk, val] -- r1)[rk0] = (st1 -- r1)[rk0] ⊦ ∅[r1 -- sk0] = (st1 -- r1)[r1 -- sk0], r1 -- sk0 ∈ st[r -- sk, val] -- r1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]