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Proof step 43 from proof hasSharing-put-statfield (previous step)

⊦ (∀ sks, sks0. ¬ fits(r, sks, st[jvmref -- sk, val]) ∨ ¬ fits(r, sks0, st[jvmref -- sk, val]) ∨ sks = sks0 ∨ sks = [] ∨ sks0 = [] ∨ (st[jvmref -- sk, val][r -- sks]) .refval = jvmref ∨ (st[jvmref -- sk, val][r -- sks]) ≠ (st[jvmref -- sk, val][r -- sks0])) ↔ ¬ (∃ sks1, sks2. fits(r, sks1, st) ∧ fits(r, sks2, st) ∧ sks1 ≠ sks2 ∧ sks1 ≠ [] ∧ sks2 ≠ [] ∧ st[r -- sks1] .refval ≠ jvmref ∧ st[r -- sks1] = st[r -- sks2]), junkref(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
⊦ (∀ sks, sks0. ¬ fits(r, sks, st[jvmref -- sk, val]) ∨ ¬ fits(r, sks0, st[jvmref -- sk, val]) ∨ sks = sks0 ∨ sks = [] ∨ sks0 = [] ∨ (st[jvmref -- sk, val][r -- sks]) .refval = jvmref ∨ (st[jvmref -- sk, val][r -- sks]) ≠ (st[jvmref -- sk, val][r -- sks0])) ↔ (∀ sks, sks0. ¬ fits(r, sks, st) ∨ ¬ fits(r, sks0, st) ∨ sks = sks0 ∨ sks = [] ∨ sks0 = [] ∨ st[r -- sks] .refval = jvmref ∨ st[r -- sks] ≠ st[r -- sks0]), junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]