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Proof step 38 from proof eqex-tput (previous step)

eqex(st1, st0[r -- sk, val], r0), eqex(st0[r -- sk, val], st1, r0), eqex(st, st0, r0), eqex(st0, st, r0), ∀ rk. (rk ∈ st -- r0 ↔ rk ∈ st0 -- r0) ∧ (st -- r0)[rk] = (st0 -- r0)[rk], ∀ rk0. (rk0 ∈ st0[r -- sk, val] -- r0 ↔ rk0 ∈ st1 -- r0) ∧ (st0[r -- sk, val] -- r0)[rk0] = (st1 -- r0)[rk0] ⊦ r0 = r1, r1 -- sk0 ∈ st ∨ r1 = r ∧ sk0 = sk ↔ r1 -- sk0 ∈ st1

User applied rule all left

And obtained the following new goals

Proof step 37:
eqex(st1, st0[r -- sk, val], r0), eqex(st0[r -- sk, val], st1, r0), eqex(st, st0, r0), eqex(st0, st, r0), ∀ rk. (rk ∈ st -- r0 ↔ rk ∈ st0 -- r0) ∧ (st -- r0)[rk] = (st0 -- r0)[rk], (r1 -- sk0 ∈ st0[r -- sk, val] -- r0 ↔ r1 -- sk0 ∈ st1 -- r0) ∧ (st0[r -- sk, val] -- r0)[r1 -- sk0] = (st1 -- r0)[r1 -- sk0], ∀ rk0. (rk0 ∈ st0[r -- sk, val] -- r0 ↔ rk0 ∈ st1 -- r0) ∧ (st0[r -- sk, val] -- r0)[rk0] = (st1 -- r0)[rk0] ⊦ r0 = r1, r1 -- sk0 ∈ st ∨ r1 = r ∧ sk0 = sk ↔ r1 -- sk0 ∈ st1

Rule argument

[EXRARG : pos: left 6, [extquanttermlist: (r1 -- sk0) abort:F , discard:F , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r1 -- sk0)]]