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Proof step 5 from proof del-put-switch (previous step)

⊦ r1 -- sk0 ∈ st[r -- sk, val] -- r0 ↔ r1 -- sk0 = r -- sk ∨ r1 -- sk0 ∈ st -- r0, r = r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st
Proof step 3:
⊦ r -- sk ∈ st -- r0 ↔ ¬ ¬ (r ≠ r0 ∧ r -- sk ∈ st)
Proof step 4:
r1 = r ∧ sk0 = sk ∨ r1 ≠ r0 ∧ r1 -- sk0 ∈ st, r1 = r0 ∨ (r1 ≠ r ∨ sk0 ≠ sk) ∧ ¬ r1 -- sk0 ∈ st ⊦ r = r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])]]