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Proof step 6 from proof map-same (previous step)

rs1 ∩ rs0 = ∅, a[r => o][rs1, f, g], (a0[r => o0] * R)[rs0, f, g] ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 7:
rs1 ∩ rs0 = ∅, (a0[r => o0] * R)[rs0, f, g], a[r => o][rs1, f, g], r ∈ rs1, r ∈ rs0 ⊦

Rule argument

fma: r ∈ rs1 ∧ r ∈ rs0

Additional information

Used simplifier rules: map-in0: ⊦ a[r => o][rs, f, g] → r ∈ rs;%map-in: ⊦ (a[r => o] * R)[rs, f, g] → r ∈ rs;%sep-ass: ⊦ (R0 * R1) * R2 = R0 * R1 * R2; from specification heap-sep[heap-aux]%sep-com: ⊦ R0 * R1 = R1 * R0; from specification heap-sep[heap-aux] Used formulas: r ∈ rs1 ∧ r ∈ rs0